<--• ' 





Class QA^ 
Book __ 



CopyiigM W 

COPXMGHT DEPOSE 



UNIFIED 
MATHEMATICS 



BY 

LOUIS d* KARPINSKI, Ph.D. 

PROFESSOR OF MATHEMATICS 
UNIVERSITY. OF MICHIGAN 

HARRY y/bENEDICT, Ph.D. 

PROFESSOR OF APPLIED MATHEMATICS 
UNIVERSITY OF TEXAS 



JOHN W. CALHOUN, M.A. 

ASSOCIATE PROFESSOR OF PURE MATHEMATICS 
UNIVERSITY OF TEXAS 



REVISED 



D. C. HEATH & CO., PUBLISHERS 
BOSTON NEW YORK CHICAGO 



a? 



Copyright, 1918 and 1922, 
By D. C. Heath & Co. 

2c 2 



mivn 



©CU661454 



PREFACE 

This text presents a course in elementary mathematics adapted to 
the needs of students in the freshman year of an ordinary college or 
technical school course, and of students in the first year of a junior 
college. The material of the text includes the essential and. vital 
features of the work commonly covered in the past in separate courses 
in college algebra, trigonometry, and analytical geometry. 

The fundamental idea of the development is to emphasize the fact 
that mathematics cannot be artificially divided into compartments 
with separate labels, as we have been in the habit of doing, and to 
show the essential unity and harmony and interplay between the two 
great fields into which mathematics may properly be divided; viz., 
analysis and geometry. 

A further fundamental feature of this work is the insistence upon 
illustrations drawn from fields with which the ordinary student has 
real experience. The authors believe that an illustration taken from 
life adds to the cultural value of the course in mathematics in which 
this illustration is discussed. Mathematics is essentially a mental 
discipline, but it is also a powerful tool of science, playing a won- 
derful part in the development of civilization. Both of these facts 
are continually emphasized in this text and from different points of 
approach. 

The student who has in any sense mastered the material which is 
presented will at the same time, and without great effort, have 
acquired a real appreciation of the mathematical problems of physics, 
of engineering, of the science of statistics, and of science in general. 

A distinctly new feature of the work is the introduction of series of 
" timing exercises " in types of problems in which the student may be 
expected to develop an almost mechanical ability. The time which 
is given in the problems is wholly tentative ; it is hoped, in the 
interest of definite and scientific knowledge concerning what may be 
expected of a freshman, that institutions using this text will keep a 
somewhat detailed record of the time actually made by groups of 
their students. The authors invite the cooperation of teachers of ele- 



iv PREFACE 

mentary college mathematics in the attempt to secure this valuable 
information. The authors will make every effort to put information 
thus secured at the service of the public interested. 

In general, the diagrams are carefully drawn on paper with sub- 
divisions of twentieths of an inch. It is expected that this kind of 
paper will be used as far as possible in the graphical work, as students 
will be found to acquire rapidly the ability to use intelligently this 
type of coordinate paper. Considerable attention should be paid by 
the teacher to the intelligent reading and interpretation of the dia- 
grams which appear in the text, as the student will in this way gain 
power to handle his own diagrams, and appreciation of the vital 
importance of the method. The photographic illustrations should 
also be used in a somewhat similar manner. 

The material can be covered without systematic omissions in a 
course which devotes five hours per week for one year to the study of 
mathematics. In a four-hour course there are certain omissions 
which can be made by the teacher at his own discretion ; the three 
chapters on solid analytical geometry are not commonly presented in 
the ordinary four-hour course; the chapter on "Poles and Polars" 
may also be omitted. The exercises are so numerous that any teacher 
can make a selection, which can be varied, if desired, in succeeding 
years. 

No attempt has been made to introduce the terminology of the 
calculus as it is found that there is ample material in the more ele- 
mentary field which should be covered before the student embarks 
upon what may properly be called higher mathematics. However, 
the fundamental idea of the derivative is presented and utilized with- 
out the new terminology. 

The authors are greatly indebted to a large number of their col- 
leagues who have been most generous in furnishing real illustrations 
in various fields. Professor N". H. Williams of the Department of 
Physics at the University of Michigan has given very pertinent and 
valuable comment on numerous sections, in addition to furnishing 
the beautiful oscillograms of alternating currents. Professor W. J. 
Hussey of the Detroit Observatory furnished the temperature and ba- 
rometer chart, and has given generously of his time in the discussion of 
astronomical problems adapted to an elementary text. Professors J. 
J. Cox, H. E. Riggs, A. F. Greiner, H. H. Higbie, J. C. Parker, Leon 
J. Makielski, E. M. Bragg, H. W. King, and L. M. Gram of the De- 
partment of Engineering, University of Michigan, have giveu valu- 






PREFACE V 

able advice and suggestions. The diagram illustrating the use of the 
ellipse in determining the proper amounts of sand and gravel to use 
from given pits to obtain the best results was furnished by Professor 
Cox. To Professor Greiner we are indebted for the cut of the six 
cylinders of an automobile engine, and for criticising the piston- 
rod motion. To Mr. Makielski, the well-known artist, we are 
indebted for the drawing of a box which is reproduced. Professor 
James W. Glover of the Actuarial and Statistical Department, Uni- 
versity of Michigan, has read and corrected the material relating to 
his field. To Professor C. L. Meader of the Department of Lin- 
guistics, and to Professors Pillsbury and Shepard of the Department 
of Psychology, University of Michigan, we are indebted for the tuning- 
fork records and for the vowel and consonant records. To Professor 
F. G. Xovy of the Hygienic Laboratory we are indebted for certain 
information concerning bacterial growth. Captain Peter Field, Coast 
Artillery, U.S.A., has indicated to us certain simple problems con- 
nected with artillery work. To Mr. H. J. Karpinski we owe the 
photographs of the Rialto and the Colosseum. To the Albert Kahn 
Company of Detroit we are indebted for information concerning de- 
tails of the Hill Auditorium, and to the Tyrrell Engineering Company 
of Detroit for permission to reproduce a number of photographs of 
bridges. Vfe render to these gentlemen and to our colleagues who 
have been generous in giving time and thought to our inquiries our 
sincere appreciation for their friendly cooperation. In every field 
which we touch, we assume full responsibility for all errors, and we 
shall be grateful to teachers who will assist in removing the in- 
evitable blemishes in a book of this size and character. 

The proof has been carefully read by Professor E. V. Huntington 
of Harvard L^niversity and by Professor C. N". Moore of the University 
of Cincinnati. Many blemishes have been removed and. many impor- 
tant additions and changes have been made on their advice. Pro- 
fessor J. W. Bradshaw of the University of Michigan and Professor 
J. D. Bond of the Texas Agricultural College have read the galley 
proof and given numerous and excellent suggestions. Professor 
W. W. Beman of the University of Michigan has read the page proof 
and has made numerous vital corrections and suggestions. Professor 
J. L. Markley has given advice on the early chapters. To all of 
these gentlemen we acknowledge our real indebtedness. 

In putting the work through the press, the responsible editorship 
has been placed in the hands of Professor Karpinski, as the exigencies 



vi PREFACE 

of time and space — Texas to Michigan to New York to Boston — 
would have delayed the book for a full year with a divided responsi- 
bility. Certain chapters, including the chapter on the applications 
of the conic sections, the chapters on the sine curve, on the growth 
curve and on complex numbers, the treatment of solid analytics, 
the tables and most of the problems, are due entirely to Professor 
Karpinski. 

The drawings have been made at the University of Michigan, chiefly 
by Mr. E. T. Cranch, an engineer now in the service, U.S.A. Most 
of the photographs are by Miss F. J. Dunbar of the University of 
Michigan Lantern Slide Shop, and a few are by Mr. G. R. Swain. 



CONTENTS 



CHAPTER 
I. 

II. 

III. 

IV. 
V. 

VI. 

VII. 

VIII. 

IX. 

X. 

XI. 

XII. 

XIII. 

XIV. 

XV. 

XVI. 

XVII. 

XVIII. 

XIX. 

XX. 

XXI. 

XXII. 

XXIII. 



Numbers of Algebra 1 

Application of Algebra to Arithmetic . . 18 

Exponents and Logarithms 40 

Graphical Representation of Functions . . 58 
The Linear and Quadratic Functions of One 

Variable 77 

Straight-Line and Two-Point Formulas . .101 

Trigonometric Functions 110 

Tables and Applications 139 

Applications of Trigonometric Functions . . 149 
Arithmetical Series and Arithmetical Interpo- 
lation 166 

Geometrical Series and Applications to Annu- 
ities 176 

Binomial Series and Applications . . .193 

Right Triangles 206 

The Circle 220 

Addition Formulas 237 

Trigonometric Formulas for Oblique Lines . 251 

Solution of Triangles . . . . . . 266 

The Ellipse 281 

The Parabola 309 

The Hyperbola 320 

Taxgents and Normals to Second Degree Curves 332 

Applications of Conic Sections .... 346 

Poles, Polars, and Diameters .... 363 
vii 



Vlll 



CONTENTS 



CHAPTER 

XXIV. 

XXV. 

XXVI. 

XXVII. 

XXVIII. 

XXIX. 

XXX. 

XXXI. 

XXXII. 



PAGE 

Algebraic Transformations and Substitutions 371 

Solution of Numerical Algebraic Equations . 392 

Wave Motion 407 

Laws of Growth 423 

Polar Coordinates 435 

Complex Numbers 439 

Solid Analytic Geometry: Points and Lines 452 
Solid Analytics : First Degree Equations and 

Equations in Two Variables .... 465 

Solid Analytics : Quadric Surfaces . . . 475 

Tables ......... 495 



UNIFIED MATHEMATICS 

CHAPTER I 
NUMBERS OF ALGEBRA 

-5 -4 -3 -2 -1 +1 +2 -+3 +4 45 

1 1— 1 1 1 1 J 1— 1 1 1 

1. Representation of points on a line. — With, any given unit 
of length and a fixed point of reference, called the origin, the 
points upon a given line are located by numbers. The unit of 
length and parts thereof are laid off in both directions from the 
origin to locate further points. To each point corresponds one 
number (a symbol) and only one, and to each number corre- 
sponds one and only one point. We call this a one-to-one corre- 
spondence. To any point upon the line of reference corresponds 
evidently another point symmetrically placed with respect to 
the origin. This symmetry is indicated in the symbols by 
using the same set of symbols twice, distinguishing by two 
" quality" signs + and — . All points on one side of have the 
-+- sign prefixed to the symbols designating them, while the 
corresponding points on the other side take the same symbols, 
prefixing the negative sign. Thus -\-a and —a represent sym- 
metrically placed points on the scalar line. The line of refer- 
ence is now called a directed line. Of two numbers represented 
by points on this line, the one represented by that one of the 
two points which lies to the right hand is called the greater. 
Such a line is the line on an ordinary thermometer ; to each 
number, then, there corresponds further a certain temper- 
ature. Thousands of physical and material interpretations of 

1 



2 UNIFIED MATHEMATICS 

the points upon such a line and the corresponding numbers 
are possible. 

This scalar line, as the above is termed, is not necessarily a 
straight line. Thus the equator is a scalar line as it is repre- 
sented upon any globe, with the zero at the intersection with 
the meridian of Greenwich, and distances given in degrees, 
each representing -g-i-^- part of the equatorial circumference of 
the earth ; -+- and — are represented on this line by E. and W. 

PROBLEMS 

1. Interpret a scalar line as representing distance upon the 
main line of the Michigan Central Railroad from Detroit, 
east and west. 

2. What is the significance of points to the left of the origin 
when the line represents your bank account ? 

3. Interpret the line as representing weights. 

4. Interpret the line as the prime meridian. What length 
is represented by 1° (circumference of earth is 25,000 miles) ? 

5. Interpret the scale as representing percentage of fat in 
foods. 

6. Represent the Fahrenheit scale on one side of such a 
line and the Centigrade upon the other, making the zero and 
the 100° of the Centigrade scale fall upon the 32° and 212° of 
the Fahrenheit ; note that for a convenient total length 20° 
Fahrenheit may be taken as corresponding to 1 centimeter or 
to one half an inch. 

2. Real numbers ; positive integers. — The symbols repre- 
senting the points upon a line, as above, are called real num- 
bers. Elementary algebra is largely a study of such numbers, 
combined according to certain rules. The rules of the game 
of algebra, as we may term it, can be studied entirely apart 
from any physical application, but the study is of funda- 
mental importance because of the part which algebraic num- 
bers play in the sciences. However, a knowledge of the laws 



NUMBERS OF ALGEBRA 3 

of algebra, apart from the applications, is necessary to enable 
one to apply the numbers effectively to physical problems. 

The real numbers are sub-divided into positive and negative 
numbers ; another classification is into rational and irrational 
numbers, the rational numbers being further sub-divided into 
integers and fractions. Numbers are represented by the letters 
a, b, c, ••• x, y, z, etc. 

Integers were undoubtedly conceived long before man began 
to write. The idea of an integer involves the notion of a 
group of individual objects, and of one-to-one correspondence. 
The idea or notion which is common to all groups of objects 
which can be placed in one-to-one correspondence with the 
objects of a given group is called the number of the given 
group of objects. Thus the pennies OOOOO can be placed 
in one-to-one correspondence with some segments of our line, 
or with the group of symbols which correspond to these 

OOOOO 



segments, or with the individuals of any one of infinitely 
many other groups, of number Jive, which have the one com- 
mon property that they can be placed in one-to-one correspond- 
ence with each other. The definition is recent ; the idea is 
old. One-to-one correspondence appears frequently in physical 
problems, as in the one-to-one correspondence between degrees 
Centigrade and degrees Fahrenheit above. 

Integers can be used to represent segments of our line of 
reference, from as reference point, with some length as 
unit of measure (or as individual of the group). The ex- 
tremity farthest from is marked with the integer corre- 
sponding to the number of the segments between that point 
and 0. Evidently certain groups of segments include as sub- 
groups other groups of segments. The number of the includ- 
ing group is called greater than the number of any included 
group ; the included group is smaller, and its number is less 
than the number of the including group. Thus the group 



4 UNIFIED MATHEMATICS 

called eight, 8, has the smaller sub-groups, 1, 2, 3, 4, 5, 6, 
and 7. 

1 1 I 2 I 3 I 4 I 5 I 6~ I 7 1 8 1 

3. Positive integers ; fundamental laws, definitions, assumptions, 
and theorems. — G-iven two positive integers, a and b, the single 
group composed of the individuals from two distinct groups 
of objects, represented by a and b respectively, is represented 
by another number x, the sum of a and b, which latter we term 
summands. The process of finding such a number is called 
addition, and is indicated by writing the sign -f between the 
two given numbers a and b. By the sum of three numbers is 
meant the number obtained by adding the third to the sum of 
the first two, and similarly for more numbers than three. The 
following are assumptions and theorems concerning integers. 

I. x = a + b, given two integers, the sum exists. 

II. a + b = b -f- a, addition is commutative, i.e. the order of 
addition is immaterial. 

III. a + b + c = (a + b) + c = a + (b + c); the associative law 
for addition. 

IV. x + b = a. Given the sum a and one of tthe summands, 
the other summand exists : x is the number which added to b 
gives a. This defines the operation of subtraction, which is 
represented by the sign — , to be placed between the sum and 
the given summand, as in a — b = x. 

By definition, (a — b) + b = a, and for the present this has 
meaning only when b is less than a ; a is termed minuend, b is 
termed subtrahend, and a — b is the remainder. 

Thus, given x + 2 = 7, x is evidently 5, as one remembers that in the 
operation of addition 5 added to 2 gives 7. Given x + 2 = 2, or x + 2 = 1, 
we have, at this stage of development, no number x which satisfies the 
given condition. 

V. If a 4- c = b + c, a = b, and conversely. 

The converse is equivalent to the axiom, if equals be added 
to equals the sums are equal, 



NUMBERS OF ALGEBRA 5 

VI. x = a ■ b. Suppose that each individual of a group of 
a objects consists of b individuals of another type, e.g. 4 rows, 
each of 7 dots, then the single group 
consisting of all the second type of 

individuals involved is called the prod- 

uct of a and b. The operation is called 

multiplication and is represented by 
the sign x between a and b, or by a 

period (slightly elevated) between a and b, or by simple juxta- 
position of the two numbers, a and b, called factors ; a is 
termed multiplier, and b is multiplicand. 

VII.- a • b = b • a, the commutative law for multiplication, 
evident from the figure. 

VIII. a - b • c = (a • b) • c = a (6 • c), the associative law for 
multiplication. 

IX. a(b -f- c) = ab + ac, multiplication is distributive with 
respect to addition. This corresponds precisely to our ordi- 
nary method of multiplication. 

X. b • x = a. Given the product a, and one factor b, x is de- 
fined by this relation as the number which multiplied by b gives 
a. This operation is limited when dealing with integers to num- 
bers a and b, which are so related that a is one of the products 
obtained by multiplying b by an integer. 

The process of finding x is called division, and is represented 

by the sign -*-, or by placing a over b, b • - = a; b is termed 

b 

the divisor, a is the dividend, and x is termed the quotient 

These laws concerning positive integers constitute simply a 

restatement of facts with which the student is familiar. The 

four fundamental operations to this point have been confined 

to the field of positive integers ; evidently the operation of 

division when b is the divisor applies only to those positive 

integers which are multiples of b. Similarly the operation of 

subtraction of b from a is limited to integers so related that 



6 UNIFIED MATHEMATICS 

a>6. We extend our field of numbers by removing these 
limitations. Thus if you wish to have a number x which 
multiplied by 8 gives 5 you do not find it among the positive 
integers ; you may then decide to create such a number, calling 
it J-, the two symbols indicating the definition and genesis of 
the new number. Such extensions of the number field are 
briefly indicated in the next section. 

4. Rational numbers ; zero, fractions, and negative integers. ->— 
These fundamental equalities and definitions from I to X are 
now extended by removing all limitations (except one, as noted" 
below) upon the numbers, a, b, c, and x. Note that only the 
operations of addition, subtraction, multiplication, and division 
are included at this point. 

Extension of IV. x + b = a, when b = a defines zero, 
written 0. By definition then, + a = a. x-\-b = 0, defines 
a negative number which is written — b. The negative here 
is a sign of quality ; by definition — b is the result of subtract- 
ing b from 0, and — b + b = 0. 

x -f- b = a, when b > a, defines the negative number a — b, 
which is the negative of b — a. 

Subtracting a negative number can now be shown to be 
equivalent to adding a positive number, and similarly the other 
rules of elementary algebra relating to the addition and sub- 
traction of positive and negative quantities. That the product 
of two numbers with like signs is positive and the product of 
two numbers with unlike signs is negative follows from the 
above development. 

A negative number — a is placed in our line of reference 
symmetrically to the corresponding positive number a, with 
respect to the origin ; of two negatives, the one toward the 
right is called the greater. 

Extension of X. b • - = a, for all values of b except 0. 
b 

This extension of - to mean a number which multiplied by 
b 



NUMBERS OF ALGEBRA 



b gives a introduces new numbers of the type - , rational frac- 
tions in which, a and b are positive or negative integers. 
Integers are included in this definition if b is a factor of a or 
if b is equal to one. Division -by is explicitly excluded. 

A rational number is any number ivliich can be expressed as 
the quotient of two integers, the denominator not to be zero. 

All of our rules for operating with fractions follow from the 
definition of - and from the preceding development. Thus 

= • = . Further, by definition, 

-b b b v J 

a c 

- > - , when both are positive if ad > be ; 
b d 

- = -, when both are positive if ad = be ; and 
b d 

a c 

- < - , when both are positive if ad < be. 
b d 



Positive fractions can thus be arranged in a determined order 
upon our line of reference ; the value of the fraction determines 

the position and a graphical method of locating - on the 

b 
scalar line is indicated in the next section ; negative fractions 
are placed symmetrically to the corresponding positive frac- 
tions, with respect to the origin. 

Of any two rational numbers a and b, a is greater than b 
(a > 6) if a — b is positive ; for when a — b is positive, a posi- 
tive length must be added to b to give a, and consequently a 
must lie to the right of b. If on the line of reference two 
points Xi and x 2 are taken (fixed points), x 2 — x l gives the dis- 
tance from the first point to the second ; this expression is 
positive if x 2 > x x , and negative if x 2 < x v 

That there is a distinction between -f- and — used as signs 
of operation, as with positive integers in the preceding section, 



8 UNIFIED MATHEMATICS 

and + and — used as quality signs is apparent. Thus — 6 
may indicate that b is to be subtracted from some preceding 
number, or — b may indicate that the distance b is taken on 
the negative side of the origin. The fact that a+(— b), the 
addition to a of negative b, gives the same result as subtract- 
ing b from a, or a — b, is readily shown by the graphical 
method of section 9 below. This type of relationship obviates 
any need for careful distinction between the two possible mean- 
ings of these signs and makes separate symbols not necessary. 
When no sign is used with a number symbol the + sign is 
understood. 

EXERCISES 

3 3 

1. Explain the distinction between ■ and ; between 

3 , -3 

■ and 

-7 7 

— 5 — a 

2. Write in the three forms corresponding to > 

3 — x b 

b' - b 

3. Is — 3 > — 2 ? Which is greater, or — ^3 ? Explain. 

4. What is the difference between 4 and — 3 ? 4 and 3 ? 
4 and 11 ? 

5. What fundamental law is assumed in the common process 
of multiplication, e. g. as in 325 by 239 and also x— 7 by x—2? 
Is there a corresponding assumption in division ? 

6. Which is greater, 1 or — § ? Is -J-J- greater or less than 
f|? Explain. 

7. What is the product of by 7 ; by - 8 ; by 3 ; by ^? If 
a product is zero, what limitation is imposed upon the factors ? 

5. Representation of a rational number, -• On cross-section 

b 

paper any rational fraction can be represented, using ruler and 
compass. Using 5 divisions to represent unity, each division 



NUMBERS OF ALGEBRA 



9 



represents i of a unit. To represent ^ one measures off 13 
units, OA, on the line of reference, and 7 units, OB, on a 
second line through the origin (for convenience, use cross-sec- 




Graphical division 

\ to ^ of 13 represented on horizontal line of reference. 

tion paper). Connect the ends, AB, and through the point U, 
one unit from on the second line, draw a line parallel to AB. 
The intersection point on the reference line represents the 

fraction ±f-. Similarly any fraction *. can be represented. 

The series of parallels to AB through the first 7 unit points 
on the vertical axis will cut off (plane geometry theorem) 7 
equal parts of 13 on the horizontal axis. 



rr^i 


TT" 


i i : i 


x:x::::::i:::::^::::::=::::: 




4— 






|=|=:::=:|====:::===::==:==:: 


E> 


2 






----- W-T\ — ----- — 


_ . ----3 

_1 1 „ a « -Pi"" 1 9 r-+L3 


1- 

— 


.__. 


-. 


— P — 1~4— ~ t — IT 
4nifffll+--^---4--- 

2 -q : , p : — j— A— ^— ~£ : 


— ^ 9 ^ 7-H j 

_7_i_ 1: x 
-:.?_:::ip -Ii:::l£:±3 


— " 







4H+4ff 1 -#-— : 


Pill ill iilllliTTIiifrfi 



Graphical division 

\ to J / expressed decimally. 



10 



UNIFIED MATHEMATICS 



On cross-section paper a somewhat better method of indicat- 
ing any quotient - is to move out on the line of reference b 

units and up 1 unit ; connecting this point with the origin 

gives a straight line which can be used to read the desired 

quotients. Thus, since OB = 7 units, BO — 1, and OA = 13, 

. - n ., . AP OA AP 13 , AT> ,13 

it follows that = , or = — , whence AP equals — . 

BO OB 17 1 7 

To obtain -^-, you find the point 8.5 units from 0, and the 
vertical distance to the oblique line represents -^—, or 1.2. 



6. Irrational numbers. x 2 = 2 is a simple and familiar illus- 
tration of a relation which is not satisfied by any rational 

number, -, with a and b integers ; geometrically, the diagonal 

of a square with side unity 
is not represented by any 
rational number. If you wish 
the length of this diagonal 
for any practical purpose, you 
use 44, or W, or 









































> 






flife 




j^yV> 4 




<&/L - 














- - i k ~ 
























1 




- \ 




-J 



141 or 17 

10 0? U1 12> 



.2 .4 .6 .8 1 1.2 1.4 
Graphical representation of V2 



TUlTTlhPTS 1 1 ^ 141 1414 14142 
iiuij.iuc.lo a, 1Q , 100> 1000' 10T00J 



or tffa or iAi^f The car- 
penter uses 1 foot 5 inches, 
or 1^- feet, in the diagonal 
for every foot of side, with 
an error of -^ of one per 
cent. The series of rational 
which can be indefinitely ex- 



tended always increasing, and the series, always decreasing, 
2> rib tto iwrb twoHp w ^ n a constantly decreasing difference 
of limit between corresponding terms, together define the 
irrational number called the square root of 2. No rational 

number satisfies the relation ; no number - is at the end of 

b 

either series, but either series determines a definite point jon 



NUMBERS OF ALGEBRA 



11 



our line, and algebraically defines our number, which we will 
call the square root of 2. 



Proof of the irrationality of V2. — Assume that s/2 = - j a 

rational fraction in lowest terms, withp and q integers. Both 
p and q cannot be even numbers, either p is odd or q is odd. 
If p is odd, squaring and clearing of fractions, 2 q 2 = p 2 ; but 
p is odd and you have an even number equal to an odd 
number. Hence p cannot be an odd number. Now assume 
that p is even and that q is odd, and further let p=2m. 
Then 2 q 2 = 4 m 2 , q 2 — 2 m 2 , and again we have an odd num- 

— P 

ber equal to an even number. Our assumption that V2 = - 

leads to an absurdity, that an odd number equals an even 
number. 

Describe about the origin with a radius 10 a circle, and 
using a protractor measure an angle of 20 degrees. The 
length of the perpendicular and 
the part cut off by the perpen- 
dicular from the end of this 
line are definite and precise 
points which can be computed 
to any degree of accuracy de- 
sired. No rational numbers rep- 
resent these lengths, which are 
trigonometric irrationalities. A 
series of constantly increasing 
rational numbers can be found, 
such that there is no greatest 
of the series, to represent lines 
which are always shorter than the given line ; and another 
series of terms constantly decreasing, but approaching to the 
terms of the first series, can be found. No largest number 
can be found in the first series and no smallest in the second ; 
both sequences together define, we may say, an irrational 
number. 



_j_ __ __ — 








, | 


































■1? p. ■ - - 




















'/ _ __. . ... . . . '" _V 





A trigonometric irrational 



12 UNIFIED MATHEMATICS 

EXERCISES 

1. Write 6 terms of the decreasing series denning V2; V& 
Arrange in order of magnitude 6 numbers with squares greater than 2. 

2. How is the series for defining the length of the circum- 
ference of a circle obtained ? What assumption is made ? 

3. Find by actual multiplication six numbers between 60 
and 63 whose squares are less than 3910. 

4. Inscribe a square and an equilateral triangle in the 
circle of radius 10 ; find the sides. 

7. Constants and variables. — Every fixed point on the line 
of reference is at a fixed distance from the point of reference, 
; the distance is constant. We can think of a point as mov- 
ing on the line OX in either direction. The distance from O 
then varies and we speak of the distance as a variable. Thus, 
also, the price of wheat during a term of years or in different 
parts of the world is a variable ; the weight of an animal at 
different ages is a variable. We think of the variable quantity 
as taking a series of values under diverse conditions. We can 
represent the variable distance of a point from on our line 
by the single letter x, which may then in the various possible 
positions on the line be thought of as positive, or negative, or 
zero, as rational or irrational. This letter x represents then a 
variable quantity and is essentially a number, subject to all 
the operations on algebraic quantities as noted above. In gen- 
eral, we designate the variable point by the single letter P; 
the distance from is OP, of which the length and direction 
from are indicated by the number or variable x. A point on 
the line X f X is represented by a single letter x, called the 
abscissa of the point. The fixed points on the line are fre- 
quently represented by the letters a, b, c, d, ••• or by x 1} x 2 , 
x 3 , — , each of which may represent any point upon the line. 

8. Historical note. — Modern algebra with the systematic 
employment of literal coefficients, letters to represent general 
constants, was introduced by the great French mathematician 



NUMBERS OF ALGEBRA 



13 



and statesman, Francois Viete (1540-1603) ; Viete used the 
consonants to represent known quantities and vowels to re- 
present unknowns, using capitals for both. The use of the 
symbols of operation in equations dates also from about the 
same time ; our equality sign was introduced by Robert 
Recorde, an English physician and mathematician, whose 
" Whetstone of Witte," 1557, is the first treatise in the Eng- 
lish language on algebra. The + and — symbols are due to 
a German, Widmann, and date from 1489. 

9. Geometrical equivalents of the four fundamental operations. 
a. Addition. — The operation of addition of x x and x 2 is repre- 
sented graphically by placing the length OP 2 upon the line 







Xf\-X 2 






o\ 






J \ 


Pz 


Xy 


-4Pl *2 


.\P 2 


P 3 




x l +x 2 




m K 




-\o Pi 


L_ 






-\ x, I 








X 2 


1 l i 





OP 3 = OP x + P X P Z = OPx + OP 2 = x x + x 2 . 

from the point P x in the direction of OP 2 . Physically, 
adaition is the result in general of two different causes. 
Thus a weight of 3 pounds + a weight of 5 pounds ; a vertical 
velocity due to the action of gravity (on a falling body) + a 
vertical velocity due to some other force ; a transportation 
(translation) from one point to another + another translation 
in the same direction ; two successive rotations of a wheel 
about its axis ; these are familiar examples of addition. 

b. Subtraction. OP l + P X P 2 = OP 2 ; P X P 2 = OP 2 — OP x . 

Whatever the relative positions on the line OX of P 1 and P 2 > 
with respect to the position of 0, OP x + PiP 2 = OP 2 , all of 
these representing directed line segments. In words, the 
equality P X P 2 = OP 2 — OP x states that the distance from any 
point Pi on a directed line to a second point on the line is 
given by the abscissa of the second point minus the abscissa of 



14 UNIFIED MATHEMATICS 

the first, with, respect to any third point 0, on the line, as 
origin. 

If we represent the distance from P x to P 2 by the letter d, 
we have x Y + d = x 2 , or d = x 2 — x±. Subtraction is represented 
by the distance from the first point to the second point, which 



DISTANCE BETWEEN TWO 


POINTS 


ON A SCALAR LINE 


O 


^=t 


±u 


" 




^ rl< ^"^i Id 


9 F 2r [Pi 




H^p, o 


| 










^k^s-lo Xl 

1 **-*' ' ilr, 


-+ 




R\ _ L a?i- 




pi M*2 


J 


x 2-*i > i 


h m — b 


=f 






^k 


1 £2 ^| 2 



OPi + PiP 2 = OP 2 ; PiP 2 =. OP 2 - OPi ; d=x 2 - Xl 
Fundamental property of any three points on a directed line 

equals OP 2 — OPY Since in physical problems the distance 
represents the change in numerical value of physical quantities, 
graphical subtraction is more frequently noted than addition. 
We say the temperature has risen 10 degrees or fallen 10 
degrees, having in mind the original and the final reading ; 
when we say that an iron bar has expanded an inch, the 
initial and the final length and the change in length are im- 
portant. 

The formula, 

d = X 2 — Xj, 

gives the distance PiP 2 from the point P lf abscissa x u to the 
point P 2 , abscissa x 2 ; the algebraic sign gives the direction. 
The formula gives then the change in value of a variable x in 
passing from the value x 1 to the value x 2 . 

c. Multiplication and division. — Graphical mu. ^plication 
and division upon cross-section paper, involving theorems con- 
cerning similar triangles, are indicated by the diagrams on 
page 9. 



NUMBERS OF ALGEBRA 15 

PROBLEMS 

1. What is the distance from 2 to 8 on a scalar line ? 

2. On a thermometer what is the change from —3° to +25° ? 
from - 10° to + 30° ? from + 30° to - 10° ? from - 10° to 

— 20° ? 

3. Use the formula, d = x> — x 1} to find the distance from 

— 2 to -4- 8 on a scalar line, noting that x 2 = 8, and x x = — 2 ; 
from - 2 to - 8 ; from - 11.3 to -f 24.7. 

4. Represent the square root of 2 geometrically, ta.king 10 
quarter-inches as 1. Represent the square root of 3 on coordi- 
nate paper, using the same scale. Represent the square root 
of 5, 6, 7, and 8. 

5. Draw a semicircle on a diameter of 10 half-inches. Note 
that the perpendicular at any point on this diameter is, by 
plane geometry, a mean proportional between the segments 
of the diameter. Read the mean proportional between 1 and 
9, as the vertical line drawn at the point on the diameter 4 
units out from the center. Read the mean proportional ]petween 
2 and 8, similarly ; between 3 and 7 ; between 4 and 6. 

6. Regard the preceding circle as having a radius 10 quar- 
ter-inches. Eind approximately, from it, Vl9, V36, V51, 
V64, V7o, V84, V91, V96, V99, and VI66. These are the 
vertical lengths at the points dividing the diameter in the 
ratio 19 to 1 ; 18 to 2 ; 17 to 3 ; - 10 to 10. 

7. Take a circle of diameter 12 half-inches ; from it ap- 
proximate VIl, V20, V27, V32, V35, and V36. Note 
that V20=2V5, V27 = 3V3, and V32 = 4V2; from these 
approximate V2, V3, and V5. See diagram on page 72. 

8. On the preceding circle check the geometrical fact that 
either si., ; of a right triangle is a mean proportional between 
the whole hypotenuse and the adjacent segment of the hypote- 
nuse cut off the perpendicular from the vertex of the right 
angle. 



CHAPTER II 

FORMULAS OF ALGEBRA AND GEOMETRY WITH 
ARITHMETICAL APPLICATIONS 

1. Algebraical formulas, products, and factors. — 

(x + a) 2 = x 2 + 2ax + a 2 . 
(x-af=x 2 -2ax + a 2 . 
(x + a)(x - a)= x 2 - a 2 . 
(x -h a)(x+b)= x\+(a + b)x + ab. 
(ax + b)(cx + d)= acx 2 + (ad + bc)x + bd. 
mx + nx = (m + ri)x. 
x*-y*=(x- y)(x 2 + xy + y 2 ). 
x* + y*=(x + y)(x 2 -xy + y 2 ). 

These formulas may readily be derived by actual multipli- 
cation. They should be read and used both from left to right 
and from right to left. In the problems drill is given on the 
use of these formulas both in obtaining products and in the 
reverse operation of factoring. 

The student would do well to state all of these formulas in 
words. 

Thus, the square of a binomial equals the square of the first 
term plus the square of the second term plus twice the product 
of the first by the second. 

PROBLEMS 

1. Expand the following : 

(3 x -a) 2 , (3v-2a) 2 , (3z + 4a) 2 , (x - 10) 2 , (10 + 2?/) 2 , 

(x + i) 2 , (aj + |) 2 , (x - |) 2 , (2 m + 3 n) 2 , (-2m- 3n) 2 . 

16 



APPLICATIONS OF ALGEBRA TO ARITHMETIC 17 

2. Perform the operations indicated : 
(3x+2a)(3x-2a); (a;_2)(a; 2 + 2a; + 4); (2a;+3)(4a; 2 -6a;+9); 

(a; + a + b)(x + a - b) ; (10 + «)(10 + 6) ; (20 + o)(20 + &). 

3. Factor the following : 



a. x 2 — 4 a; + 4. 




/• 


27x*-4:$x. 


b. 4 a; 2 — 4a; + l. 




0- 


v*-8. 


c. aj2 + aj + i. 




h. 


8 a? + 1 


A 2/ 2 -i- 




i. 


ax + a;. 


e. 8 a; 2 -32. 




J- 


ax — x. 


4. Factor the following 








a. x 2 + 8 x +15. 




f. 


Ax 2 - 9. 


6. a; 2 + 8 a; + 7. 




9- 


#2 _ 3 3 + 2. 


c. x 2 — 8 x + 7. 




h. 


3,2 -10^ + 16. 


d. 4a; 2 + 16 a; + 15. 




i. 


V 2 _ 3 v _ 4> 


e. 4 a; 2 + 16 x + 7. 




J- 


6a; 2 + 29 a; + 35. 


5. Factor : 








a. x 3 — 8 2/ 3 . 




f. 


p 2 — v 2 t 2 . 


6. x 3 — 64 xy 2 . 




9- 


x 2 -2y 2 . 


c. 27-82/ 6 . 




h. 


m 2 — 4 m?i — 21 n 2 . 


d. 8 m 3 + n 3 . 




i. 


(x + y) 2 -z 2 . 


e. 100- 49 n 2 - 




J- 


mx — my + wa; — ny. 


6. Complete the following 


expressions, involving the 


squares of binomials : 








a. (a; 2 -4a; + ) = 


= (x- 


y 




b. (f + Sy ) = 


--{y + 


y 




c. (a; 2 -fa- + ) = 


,(x- 


y 




* (*-**+ ) = 


=(*- 


y 




e. 3(x' 2 -4x+ )= 


= 3(aj 


] 


) 2 = 3a; 2 -12a;+ . 


/• 5(2/ 2 -i ? /+ )= 


= %- 




) 2 =5 2 / 2 -4 2 /+ . 


gr. 3^ + 7^+- 


= 3(* 2 + £* + )=3(* + £)«. 



18 UNIFIED MATHEMATICS 

2. Division of polynomials. — An expression of the form 

a Q x n ■+■ cLiX n ~ l + a 2 x n ~ 2 -f- ••• a n _ x x + a n 

wherein n is a positive integer, a is not zero, and a , a lf a 2 ■■■ a n 
are real numbers is called a rational integral function of degree 
n in x, or a polynomial of degree n in a?. For convenience the 
symbols f(x), Q(x), F(x), ••• are used to represent such poly- 
nomials. 

Given a polynomial in x and x — a, the process of finding a 
second polynomial which when multiplied by x — a gives the 
first, or the first with the exception of a remainder, is called 
division ; the polynomial found is called the quotient. 

Illustrative Problem. — Divide 3 x 3 — 2 x 2 + 7 x — 5 by x — 2. 

3 x 2 + 4 x + 15 
x - 2)3 x3 _ 2 z2 + 7 a _ 5 
3 q;3 - 6 x 2 

+ 4x 2 

+ 4 x 2 — 8 z 



+ 15x 

+ 15x-30 

+ 25 

3 x3 - 2 x 2 + 7 x - 5 = (x - 2) (3 x 2 + 4 x + 15> + 25. 

The three bars, =, are used to indicate that the expression is an iden- 
tity, true for all values of x and thus placing no limitation upon x. 

PROBLEMS 

In each problem express the result of the division in the 
form of an identity, as in the problem above. 

1. Divide 3 a 3 + 8 a 2 — 6x— 5by# — 1; by x + 2. 

2. Divide 2 x 3 - 8 x 2 + 3 x - 5 by x - 2 ; by x + 1. 

3. Divide x* - 2,000,000 by x - 120 ; by x - 126. 

4. Divide 3842 by 27, and express the result as a numerical 
identity. 

5. Divide 3 x 4 — 6 x 2 + 10 x — 7 by x — 1. 



APPLICATIONS OF ALGEBRA TO ARITHMETIC 19 

6. Divide x A - 10 x 2 - 8 by x 1 — 2. 

7. Divide 6 t* - 2 t + 1 by 2 t + 1. 

3. Remainder theorem and factor theorem. — When a poly- 
nomial in x is divided by x — a, the division can be continued 
until the remainder is a constant. If f(x) represents the 
polynomial, Q(x) the quotient, and R the remainder, we have 
the identity, 

f(x)==.(x-a)Q(x)+B. 

Substituting a for x in this identity, we have 
f(a) = (a-a)Q(a) + R, 
or, R=f(a). 

In words, the remainder obtained by dividing a polynomial 
in x by x — a is the same polynomial in a, i.e. the dividend 
with a substituted for x. This is the remainder theorem. 

Illustration. 

3 x3 - 2 £2 + 7 x - 5 = (x - 2) (3 x* + 4 x + 15) + 25 ; 
substituting 2 for x, we have 

3-23-2. 22 + 7 -2-5=25, 

since 2 — 2 = 0; the remainder 25 could have been obtained without 
actual division by x — 2, by simply substituting 2 for x in 
3 X 3 _ 2 x 2 + 7 x - 5. 

If /(a)= 0, a is called a root of f(x)= 0, or a zero of f(x). 
In this case the remainder when f(x) is divided by x— a 
equals zero, or x — a is a factor of /(#). 

f{x) = {x-a)Q (jr), when /(a) = 0. 
This is called the factor theorem. 

Illustrations. 

3 x 3 — 2 x 2 _|_ 7 x — 30 = 0, when x = 2; x — 2 is a factor. 
2 x 2 — 7 x + 5 = 0, when x = 1 ; x — 1 is a factor. 
2x 2 + 7x + 5 = 0, when x = — 1 ; x + lisa factor. 



20 UNIFIED MATHEMATICS 

The common ' ' check by nines ' ' may readily be proved by the re- 
mainder theorem : 

When 12,738, which may be written 

1 x 10 4 + 2 x 103 + 7 x 102 + 3 x 10 + 8, 

in powers of 10, is divided by 10 — 1, the remainder is equal to the origi- 
nal expression with 1 put for 10. Hence the remainder when 12,738 is 
divided by 9 is 1 +2 + 7+3 + 8, or 21, or 3 (since 21 divided by 9 gives 
3 as remainder, or 2 + 1). 

Thus, a x 10" + b x 10"~i + c x 10"~ 2 + ... g x 10 + ft divided by 10 
— 1 gives a + b + c + • • • + g + h as remainder. 

Division by 11, 10 + 1, gives as remainder the sum of the odd coeffi- 
cients less the sum of the even coefficients, counting from units' place ; 
a sum to 11 can, of course, be dropped as it occurs, or 11 can be added 
to make the remainder a positive number. 

PROBLEMS 

1. By substituting + 1 and — 1, respectively, in the follow- 
ing expressions determine those in which either x — 1 or x +■ 1 
is a factor ; factor where possible. 

a. 2x 2 -6x + 4. e. a- 3 - 1. 

b. 3x* + 5x + 2. /. 3z 2 -5a-2. 

c. 5a; 2 + 6a; -11. g. 2a 3 -7a*— 8a; -1. 

d. 5x 2 -5x. h. 2z 3 + 7a; 2 -4aT-9. 

2. For what value of a is # 3 — 3 x 2 — 7 x — a exactly divisible 
byaj-1? by x -2? 

3. For what value of a will ax* 3 — 7x 2 — Sx — a be exactly 
divisible by x + 1 ? 

4. Show that x n — y n is always divisible by x — y when n is 
an integer. 

5. For what integral values of n is x + y a factor of x n +y n ? 

6. Form the equation in x in which 3 and 4 are the roots, 
employing the factor theorem. 

Note. If 3 is a root, x — 3 is a factor. 

7. Form the equation whose roots are —3, 4, and 1. 



APPLICATIONS OF ALGEBRA TO ARITHMETIC 21 

8. What is the remainder when each of the following num- 
bers is divided by 9 ? by 11 ? 

a. 327. c. 951. e. 8217. g. 1001. 

b. 847. d. 3276. /. 12,321. h. 3003. 

4. Arithmetical application of algebraic formulas. — Algebraic 
formulas and methods can frequently be applied to arithmeti- 
cal problems with a great saving of labor ; practice with nu- 
merical examples is absolutely essential for success. 

The four formulas of elementary algebra which enjoy the 
widest use are undoubtedly : 

(x + a) 2 = x 2 + 2 ax -f- a 2 . 
(x — a) 2 = x 2 — 2 ax + a 2 . 
(x + a)(x — a) = x 2 — a 2 , 
(x + a)(x +b)= x 2 + (a + b)x + ab. 
(x + a)(# + b) gives a simple rule for the product of two 
" -teens," e.g. 19 x 17. 

Thus, (10+ a)(10+ b)=10 2 +(a + 6)10 + a&, or =10(10 +a + 6)+a6. 

Put into words, this formula states that the product of two 
numbers between 10 and 20 is equal to the whole of one plus 
the units of the other ; this sum is to be multiplied by 10 ; to 
this product is to be added the product of the units. 

Rule. — To find the product of hvo " -teens," add the whole 
of one to the units of the other and annex a zero ; to this number 

add the product of the units. 

19 
17 
260 
63 
323 

If x is taken as 20, 30, 40, 50, •••, the corresponding rule for 
the product of two two-place numbers having the same tens' 
digit is to add to the one of two numbers the units of the 
other; the sum is to be multiplied by the tens' digit, and a 



22 UNIFIED MATHEMATICS 

zero annexed to the product ; to this number add the product 
of the units. 

Thus, (37 x 35) = 30 x 42 + 35 = 1260 + 35 

= 1295. 

Such products are most easily found, evidently, if the two 

units' digits sum to 10. 

87 x 83 = 8 x 9 x 100 + 21 

= 7221. 
64 x 66 = 6 x 7 x 100 + 24 

= 4224. 

In mental work with numbers work from left to right, and 
not from right to left, dealing first with the numbers of greater 
significance. 

(x 4- a) 2 and \x — a) 2 are particularly useful in the computa- 
tion of squares of numbers of three places beginning with 1 

or 9. 

(10.7) 2 = 100 + 2 x 10 x .7 + .49 
= 114.49. 

(11.3) 2 = 121 + 6.6 + .09 = 127.69, 
or = 100 + 26 + 1.69 = 127.69. 

(1.57)2 = 2.25 + .21 + .0049, 
where .21 is obtained as 1.5 x .14 by the rule for the product of two 

1 (. 97) 2 = (1.00- .03) 2 = 1-. 06 + .0009 

= .9409. 
(8.70)2 = (io _ 1.3)2 = 100 - 26 + 1.69 = 75.69. 

Frequently it is more convenient to use these formulas re- 
arranged as follows : 

(x + a) 2 = x( x + a + a) + a 2 . 
(x — a) 2 = x(x — a — a) + a 2 . 
Thus, (84)2 -. 100(100 - 16 - 16) + 16 2 

= 100(84- 16) +16 2 
= 6800 + 256 = 7056. 
(25.7)2 = 20(25.7 + 5.7) + (5.7)2 
= 628 + 32.49 
= 660.49. 
(25.7)2= 25(25.7 + .7) + .49 = 25(26.4) + .49 
= 660.0 (since 25 = -if*) + .49. 



APPLICATIONS OF ALGEBRA TO ARITHMETIC 23 

The square of any number between 25 and 75 is obtained 
from (x + a) 2 , as follows : 

(50 ± a) 2 - 2500 ± 100 a + a 2 = 100 x (25 ± a) + a 2 . 
Thus, (37) 2 = 2500 - 1300 + 169 

= 1369. 

Rule. — To find the square of any number between 25 and 
75 ; find the difference between the given number and 50; add, if 
the given number is greater than 50, or subtract, if the given 
number is less than 50, this difference from 25 and annex to this 
tivo zeros. Add to this number the square of the difference. 

Thus, (65) 2 = (25 + 15) X 100 + 15 2 

= 4225. 

For numbers between 75 and 150 the squares may be ob- 
tained as 100(100 -a- a) + a 2 or (100) (100 + a + a)+ a 2 , 
noting that 100 — a or 100 + a is your given number whose 
square is sought. 

Thus, *112 2 = 12,400 + 144. 

13.7 2 = 174.00 + 3.7 2 = 174.00 + 13.20 + .49 = 187.69. 

Frequently, of course, only three or four significant figures 
are desired, and the methods mentioned give the significant 
figures first. 

(x + a)(x — a) may also be used for squares, thus : 

x 2 = (x + a) (x — a) + a 2 . 
(87) 2 = (87 + 13) (87 - 13) + 13 2 . 
(2.33) 2 =(2.33 + .17)(2.33-.17)+ .17 2 

= 5.4 + .0289. 
(41. 7) 2 = (41.7 4- 8.3) (41.7 - 8.3) + (8.3) 2 
= 50 x 33.4 + 8.3 2 
= 1670 + 68.89 
= 1738.89. 
(41. 72) 2 = 1738.89 -f (.04) (41 .7) + .0004 

= 1738.89 + 1.6684 = 1741 to units, or 1740.6 to tenths. 



24 UNIFIED MATHEMATICS 

PROBLEMS 

1. Multiply mentally 19 x 18, 17 x 15, 18 x 14. 

2. Use the rule given above to give the table of 18's from 
18 x 11 to 18 x 19. 

3. Multiply mentally 12 x 13, 36 x 34, 45 x 45, 82 x 88 
91 x 99. 

4. Multiply mentally 27 x 25, 34 x 32, 54 x 58, 92 x 98 

5. What is the product of 44 x 36 or (40 + 4) x (40 - 4) 
58 x 62, 44 x 37 or (40 + 4) x (40 - 3) ? 

6. What are the first three figures of (114) 2 , (107) 2 , (131) 2 
and (118) 2 ? Note (114) 2 is 12,800 + 14 2 , and the first three 
figures 129; in (116) 2 to 13,200 you must add (16) 2 , which in 
creases the first 13,200 to 13,400. 

7. From the preceding answers in 6 write the first three 
figures of (1.14) 2 , (.107) 2 , (1.31) 2 , (1180) 2 . 

8. Write the squares of 9.7, 88, 940, 8.7, and 9.2. 

9. Approximately how much greater is (9.71) 2 than (9.7) 2 ? 
(88.2) 2 than (88) 2 ? (941) 2 than (940) 2 ? (8.75) 2 than (8.7) 2 ? 
(9.26) 2 than(9.2) 2 ? 

Note that (88.2) 2 differs from (88) 2 first by .4 x 88, or by 
a little more than 35 units ; the .04 is usually negligible. 

10. Square 43, 47, 52, 63, and 62 by using the difference 
between these numbers and 50 according to the rule. 

11. Using the preceding answers, square 4.3, .47, .052, 630, 
and 6.2. 

Note. — Use common sense rules to determine the position of the 
decimal point. 

12. Using the formulas for (50 ± a) 2 , (x ± a) 2 , (100 ± a) 2 , 
write the following 25 squares. Time yourself on writing 
simply the answers ; the exercise should be completed in 6 
minutes. 



APPLICATIONS OF ALGEBRA TO ARITHMETIC 25 



57 2 = 


17' = 


24' = 


33' = 


62* = 


63' = 


42' = 


39' = 


52' = 


67' = 


87' = 


63' = 


59' = 


71' = 


82' = 


43' = 


1(X8 2 = 


21' = 


m 2 = 


L9 2 


98' = 


16 2 = 


92' = 


55' = 


49': 



13. Using the results of the preceding exercise, compute to 
lour significant figures the following squares, timing yourself. 



5Tl' = 


iT^ 2 = 


2^L5 2 = 


33l 2 = 


62A 2 = 


63^' = 


4^5 2 = 


39J 2 = 


5^9' = 


67l' = 


87l' = 


63^' = 


5^2' = 


7L8 2 = 


82^ 2 = 


43J) 2 = 


10.82* = 


2L4 2 = 


66\7 2 = 


L92 2 = 


9^6 2 = 


iae 2 = 


9^6 2 = 


5K3' = 


4^8' = 



16x19 


22 x24 


15x14 


23x26 


13x18 


24 x 29 


17x12 


28x28 


16x18 


36x33 



51x52 


66x64 


33x31 


88x82 


27 x24 


97x93 


24x22 


57 x53 


17x13 


79x71 



14. Employing the formula for (x-\- a)(x + b) write the 
following products ; the exercise should be completed in 6 
minutes. 

32x38 

43 x42 

46x44 

54 x 59 

82 x 87 

5. Extraction of roots. — In extraction of square root, the 
method of successive approximation should frequently be 
employed. 

Thus , V179.6 3 > 13 and < 14. 

V169 + 10.63 = 13 + a, wherein a must be a number such that 
2 a x 13 equals approximately 10.6. A glance shows that .4 x 13 equals 
5.2, which doubled gives 10.4. Hence, (13.4) 2 = 169 + 10.4 + .16. 
(13.4)2 _ 179.56, or 179.63 - .07. 
(13.4 + ay = 179.56 + 2 a x 13.4 + a 2 . 
a now is less than .01 ; hence, a 2 is less than .0001 ; a is to be a number 
of hundredths or thousandths, evidently, so that 2a x 13.4 is approxi- 
mately .07 ; a is roughly .003, slightly too large. 

(13.403) 2 = 179.56 + .0804 + .000009 
= 179.640409. 



26 UNIFIED MATHEMATICS 

The rule is commonly given to take x ± ~ as first approxi- 

L X 
mation of V# 2 ±a, wherein a is small as compared with x 2 . 
The process illustrated follows this rule, but suggests thinking 
multiplication instead of division. Thus in the square root of 
300, as V17 2 + 11, approximately \\ is to be added to 17 ; 
however it is easier to think 34 X a = 11, whence a = .3, or 
not quite .33; trying .32 (since the a 2 term is to be added) 
gives 17.32 of which the square is 

289 + 10.88 (or .32 x 34) + (.32) 2 = 299.9829. 



Similarly, V 3000 = V 3025- 25 = V55 2 -25 =55 - a, wherein 
a must be a number such that 2 x 55 x a will give approxi- 
mately 25 ; a is evidently .2 to one decimal place or .23 to 
two ; 54.77 is correct to four significant figures as given. 

6. Approximate roots. — Another method of approximating 
square root is to divide the given number by the first approxi- 
mation, then to use the arithmetic mean of the two numbers 
as a second approximation. Thus, 179.63-7-13=13.82; taking 

13 -I- 13 8^ 

■ — — — '— as the approximate root gives 13/41 as a second 

approximation. 179.63 -^ 13.41 gives 13.3952 and the average 
13.4026 is within .0001 of the correct value. 

Similarly the cube root may be obtained. Thus in 179.63, 
5 is the first approximation. 5 2 = 25 ; 179.63 -r- 25 = 7.2 
nearly. Taking the average of 5, 5, and 7.2 gives 5.7 as second 
approximation; (5.7) 2 = 32.49 ; 179.63 -- 32.49 = 5.529 ; the 
average of 5.7, 5.7, and 5.529 gives 5.643, which is correct 
within .001. 

PROBLEMS 

1. What is the approximate square root of 1.26 ? 128 ? 

2. Is the square root of 1.35 nearer to 1.16 or to 1.17 ? 
NotS, (1.16) 2 = 1,32 + .0256 ana (1.17) 2 = 1.34 + .0289. 



APPLICATIONS OF ALGEBRA TO ARITHMETIC 27 

3. Find the square roots of the following numbers, employ- 
ing the approximation x ± — , as the square root of x 2 ± a : 

2x 



a. 102. 


d. 83. 


g. 51. 


j. 173. 


b. 108. 


e. 80. 


h. 130. 


k. 200. 


c. 125. 


/• 48. 


i. 146. 


1. 230. 



4. By successive approximations find Vl.26 to four decimal 
places and compare with ordinary process of extraction of 
root. 

Hint. — Use 1.12 as first approximation. 

5. Find the approximation to one decimal place of the 
square roots of 65, 63, 8.30, 8.76, and 27.32. 

Hint. — Regard 8.30 as (3 — x) 2 , whence x must be roughly .12. 

6. Write the square roots of the following numbers, correct 
to 2 decimal places. Time yourself. 



9.9 


35 


65 


140 


200 


16.8 


34 


68 


150 


300 


17.2 


37.2 


78 


125 


10.4 


25.8 


39.4 , 


85 


108 


20.8 


28 


48 


90 


112 


30.6 



7. What is the remainder when x 2 — 102 is divided by 
a; -10? by x- 10.1? 

8. What is the remainder when # 3 — 1060 is divided by 
x - 10 ? by x - 10.2 ? by x - 10.3 ? 

9. Expand (x + ft) 3 ; if h is small as compared with x, how 
do the four quantities involved in (x + 7i) 3 compare in value ? 
What would be an approximate value for the cube root of 
x 3 + a, wherein a is small compared with x 3 ? 

10. Find the approximate cube root of 1060 ; similarly the 
approximate cube root of 940. 



28 UNIFIED MATHEMATICS 

7. Percentage of error. — When any measurement of physi- 
cal objects is given, the measurement has a certain limit of 
accuracy, determined in part by the instruments and methods 
of measurement and in part by the very nature of the thing 
measured. In measuring the distance to the sun from the 
earth, at some fixed time, the measurement may be given as 
93,000,000 miles ± 1,000,000 miles, or 93,000,000 miles, within 
a million miles ; the thickness of a watch spring may be meas- 
ured as .014 inch with a possible error of one thousandth of 
an inch, or .014 ± .001 inch. However, from the point of 
view of the physicist and mathematician, the distance to the 
sun is more accurately given than the thickness of the watch 
spring, for the percentage of error — ratio of possible error to 
measured value — in the case of the sun's distance is slightly 
more than 1 % of the distance, while in the other case it is 
more than 7 % of the thickness of the spring. 

Every number which represents a measurement involves 
this type of error. Obviously, in any computations with such 
numbers, results are significant only within limits determined 
by the percentage of error. 

8. Significant figures. — The significant figures in any num- 
ber representing a measurement are those which are given by 
the measurement, and do not include those initial or terminal 
zeros which are determined by the unit in which the measure- 
ment is made. The terminal zeros in 93,000,000 are not sig- 
nificant figures, as the unit of measurement here is evidently a 
million miles ; as the measurement can be made to one further 
place, the distance may be written, in " standard form," 9.3 x 
10 7 miles or 9.30 x 10 7 miles in which -only significant digits 
appear in the first factor combined with powers of 10. In the 
thickness .014 inch, the initial zero is not a significant figure, as 
it is apparent that the measurements are made in thousandths 
of an inch ; in " standard form," this is 1.4 X 10~ 2 inches. 

9. Measurement computations. Products. — If the length of 
a rectangle is measured with an error of less than 1 % of its 



APPLICATIONS OF ALGEBRA TO ARITHMETIC 29 



true value, and if the breadth is given absolutely, the true 
area will be given with the same percentage of error as the 
length. But if the breadth is also only approximately meas- 
ured, the possible error in the area obtained as . the product 

will be greater than if only one 
factor involves a possible error. 
The graph represents the right- 
hand end of a rectangle whose 
length and breadth are measured 
as 21.5 cm. and 12.2 cm. respec- 
tively, where it is understood 
that the measurement only pre- 
tends to give these dimensions 
to within one millimeter, one 
tenth of one centimeter. The 
meaning of these figures then is 
that the length lies between 
21.4 cm. and 21.6 cm., and the 
breadth between 12.1 cm. and 
12.3 cm. The first is an error of 
less than \ of 1 °/ and the sec- 
ond of less than 1 %. The un- 
certainty in area due to the pos- 
sible error in length is indicated 
by the areas at the right end 
with dimensions .1 cm. by 12.1 
cm. or .1 cm. by 12.3 cm. ; the 
uncertainty in area due to the 
breadth measurement is of di- 
mensions .1 cm. by 21.4 cm. or .1 cm. by 21.6 cm. The area 
uncertainty is then at most .1 cm. by (21.6 + 12.2) cm. or 3.39 
sq. cm. of area. This error may evidently affect the third 
figure in our computation of the area and hence in the product 
the figures beyond the third place are not significant, and give 
no real information concerning the actual area in question. 
Note that the area as the product of 12.2 by 21.5 is 262.30 







I'M 1 


1 1 1 1 1 II- 






1 


I i l I Mil 










"~'~"^\ "^tfig 


















































...... 1 1 










' ■ : ■ • i ! 


/ 






.... 


I 


























1U 










1 1 j 1 






















L 








\ 








/ 








f 








\ 








1 . 
















' . : . i 9 










1 1 M 1 ; 








































i 










'I'll | 










; 










III,. 




















|li D 





















































V 




































A 










: 1 1 4 




















: : i . 1 i i 




















o 




















. : 1 I 




















■ ■ , 1 1 






























■ ; ' 












' 1— 1 




















1 











































10 



17 18 19 20 21 22 

Measurement of an area 

A rectangle measured as 21.5 
cm. by 12.2 cm. 



30 UNIFIED MATHEMATICS 

sq. cm., but the inaccuracy of measurement of the length 
means that there is an uncertainty of area at the right-hand 
end amounting to ± 1.22 sq. cm. (cm 2 .), and similarly at the 
top an uncertainty of ± 2.16 cm 2 . ; the total uncertainty of 
area amounts to more than 3 cm 2 ., and should be given as 
± 3.38 cm 2 ., or ± 3.4 cm 2 . Commonly, of course, the measure- 
ments 12.2 cm. and 21.5 cm. mean that the area has been 
measured to one half of the last unit given ; thus this area 
actually falls between rectangles of dimensions 12.25 by 21.55 
and 12.15 by 21.45 ; even in this case the area uncertainty 
is greater, by precisely similar reasoning, than ± 1.5 cm 2 ., 
and is approximately 1.7 cm 2 . To give 262.30 as the area 
of this measured rectangle is giving nonsense in the last two 
places ; it should be given as 262 or 262.3 ± 1.7 cm 2 . 

Let k and k' represent measured quantities given with pos- 
sible errors of i % and e % respectively, e and % being assumed 
as smaller than unity (in common practice) ; the absolute 

values of these measured quantities lie 'between k(l-\ — — ) 
and kfl--*—) and between k f fl-\--^-\ and fc'fl- — 

V iooy v iooy v 10 ° 

The true product lies, then, between kk'( 1 + e + 



100 10000 

f i -4- p i& \ 

and kk'l 1 — -\ — ); in other words, the true product 

V 100 10000/ r 

may vary by *_+-£ from the computed product ; — - — - is 

disregarded, if i and e are less than 1, since the fraction is less 

than l°f of 1 % of kk'. In the graph the product xkk' 

is represented by one of the small corner squares with dimen- 
sions .1 cm. by .1 cm. 



Illustrative example. — The product of 987 by 163 wherein each num- 
ber is correct to within \ a unit need be computed only to the fourth 
significant figure as the percentage error may be as great as ^V of 1 °fo + T % 
of 1 fo, since \ in 987 parts is approximately 20W or 2V of 1 %» and \ in 



APPLICATIONS OF ALGEBRA TO ARITHMETIC 31 

The error in the product may be 
1 % of any number 
certainly affects the fourth place and probably affects the third place in 
the number. Hence there is no point whatever in carrying this compu- 
tation beyond four places, 



987 


987 


163 


163 


98700 


987 begin with 100 x 987. 


59220 


592 take 6 x 98, carrying however the 4 from 6x7. 


2961 


29 take 3x9, carrying the 2 of 3 x 8. 


160881 


160800 


163 

987 

1467 

130 

11 


987 + £ 987 - | 
163 + i 163-1 
160881 + 1(987 + 163) + \ 160, 881 - 1(987 + 163) + \ 
= 161456^ = 160306^ 



160800 ans. 

The product of 987 x 163 is 160,881 ; 987| x 163* gives 161,4561 ; 
986| x 1621 gives 160,3061 ; the actual area, if these represent dimensions 
of a rectangle measured to three significant figures lies between 160,306| 
and 161,4561. In practice we take the product 987 x 163 to four signifi- 
cant figures, which gives the area slightly more accurately than our 
measurements justify. 

10. Abbreviated multiplication of decimals. — The abbreviated 
process of multiplication applies particularly well to decimal 
fractions, but the method can be extended to integers quite as 
well. To find .9873 x .1346 correct to four decimal places. 

.9873 
.1846 
9873 begin with the highest digit of the multiplier ; first x fourth 

decimal place gives fifth decimal place. 
7898 continue with 8x7 (second x third place), carrying the 2 from 

8 x 3. 
395 take 4x8 (third x second place) carrying 3 from 4 x 7, or 28, 

which is more than 2 units in the fifth place. 

59 6 x 9, or 54, + 5 carried from the 6x8. 

18225 It assists in the process to cross out the last upper digit as it is 

used ; thus here 3 would be crossed out first, then 7, then 8, and 

finaDy 9. 



32 UNIFIED MATHEMATICS 

If a check is desired, multiply again, reversing the order of the 
factors ; thus : 

.1846 
.9873 

16614 begin with 9x6. 

1477 take 8x4, adding 5 from the 8x6 product. 

129 take 7x8, adding 3 from the 7 x 4 product. 

5 take 3x1, adding 2 from the 3x8 product. 

.18225 Read this as .1823 to four places. 

Obviously, if these were integers, you could proceed in the same way, 
writing the final product with four zeros, as 18,230,000. 



A similar abbreviation can be effected in division by drop- 
ping each time the last figure of the divisor used, and using 
the remaining part of the original divisor as new divisor. 

Thus, to divide .18225 by .9873 or by .1846 you proceed as 
follows : 

.1846 .9873 



.98/3) -18225 .1846). 18225 

9873 16614 

8352 1611 

7898 1477 

454 " 134 

395 129 

59 5 

59_ 5 

Here 9873 is used as the first divisor ; then 987 is used, but to the partial 
product, 8 x 987, is added the tens' digit of 8 x 3, the digit just crossed 
out ; then 98 is taken as divisor and to the product is added the tens' digit 

of 4 x 7 (28 is taken as giving a tens' digit of 3) ; then 9 is used and 5 
carried over from 6x8. 



11. Percentage effect of errors in divisor. — If a divisor is 
known to be too large or too small by a definite percentage oft 
itself, the quotient will be respectively smaller or larger than 
the correct quotient, for small per cents, by approximately the 
same per cent. 



APPLICATIONS OF ALGEBRA TO ARITHMETIC 33 



Bv division, — = 1 — i + i 2 — P -f- * 4 — •••• 

1 + i 



1- i 



For values of i less than .05, i 2 is less than .0025, or \ of 
1%; t», i 4 , and / 5 are less than .000125, .00000625, and 
.0000003125, respectively. Hence an error of from 1 % to 
5 % of excess in the divisor means an error of deficiency vary- 
ing also from 1 % to 5 %, within \ of 1 %, or from .99 % to 
4.75%, or from .9901% to 4.7625% in the quotient. For 
values of i between ± of 1 % and 1 %, an error of deficiency 
in the divisor means the same error of excess in the quotient, 
within -j-J-^y of this error. The meaning in physical measure- 
ments of these results is that when the divisor is correct only 
to the third significant figure, with a possible error of -J- to 1 
unit in the third place, the quotient will be correct to about 
the same degree of accuracy. 

For three-place numbers the divisor may vary from 100 to 
999. The possible error of i unit, ± \, means that 100 must 
be replaced by (100 ± .5) or 100(1 ± .005) and 999 by 999 ± .5, 
or approximately 999(1 + .0005) ; the quotient will vary from 
1 — .005 to 1 — .0005 times the obtained quotient. Hence the 
quotient obtained is valuable at most to the fourth place, and 
frequently not beyond the third place. 

Illustration. — Given that 76,430 is divided by 180; what variation 

*.„ *c* n ,™ .^, n in the quotient would a change of 1 in the divisor 
76,430 -4- 180 = 424.6. _ „ ° 

produce ? 

Suppose that instead of 180, 179 should have been used. What is the 

error in the quotient ? 180 — 1 = 180(1 — .006), the error in the divisor 

is more than .5 % and less than .6 °Jo of the divisor ; the error in the 

quotient is no more than 2.5 and no less than 2.1, since 1 % of 424.6 is 

4.246 and .6 % and .5 % are respectively 2.5 and 2.1 ; the quotient may 

be taken as 427.1, whereas 426.9 is obtained by actual division. Even an 

error of \ a unit in the divisor 180 affects the third place in the quotient. 

In obtaining .5 °fo and .6 % of 424.6, there is no point in carrying the 

work beyond two places ; the values show that the error is between 2.1 



34 UNIFIED MATHEMATICS 

and 2.5, and further places add nothing to the accuracy. The fraction 
•jlo, or in the case of the \ unit error of 3^, might just as well be used 
as per cents. This gives in the latter case -^ of 424.6, or 4- 1.2 as cor- 
rection, giving 425.8 as quotient ; the actual quotient is 425.794. 

Do not carry divisions and multiplications beyond the degree of accu- 
racy warranted by the data. 

Illustrative examples. — You can multiply 3.14159 by 140.8 and 
obtain the result numerically correct to six decimal places. But if the 
140.8 represents the diameter of a circle, measured correctly to the tenth 
of an inch (or of a foot, or of a meter) the product of 140.8 by 3.14159 
gives a valuable result only to the first decimal place ; the circumference 
cannot be computed correctly to any further percentage of accuracy 
than that with which the diameter is measured. The area can be com- 
puted here with any meaning only to four significant figures ; in fact an 
error of ± .05 inch in the diameter makes a possible error of ±10 
square inches in the area. It is convenient to write the products from 
left to right, dropping work beyond the second decimal place. 



140.8 




140.8 


H 




3.14159 the .00059 is of no use as it 


422 .4 




422.4 does not figure in the product. 


20.1 




14.1 


442.5 circumference 


5.6 
1 
442.2 circumference * 






For most practical 


purposes 


3^ is sufficiently accurate, as in finding 


Lis area, irr 2 : 




70.4 
70.4 
4928 
28 
4956 only 4 places to be retained. 


4956 




4956 


3j 




3.14159 


14868 




14868 


708 




496 


15576 area. 




198 



15567 area. 
Area as found to correspond to data, 15570. 



APPLICATIONS OF ALGEBRA TO ARITHMETIC 35 

PROBLEMS 

1. The distance of the earth from the sun varies between 
91.4 X 10 6 miles and 94.4 x 10 6 miles. The length of the 
earth's orbit lies between circles having these lengths as radii. 
Between what values does this orbit lie ? What is the ap- 
proximate orbital speed of the earth in miles per hour ? 

2. The mean distance of the earth from the sun is 
92.9 x 10 6 miles. Compute the circumference and the mean 
speed and compare by percentages with the preceding. 

3. Compute the speed of a point on the earth due to the 
rotation, taking that at latitude 45° the radius of the circle of 
latitude is 3050 miles. Compare the rotational speed with the 
revolutional speed. 

4. What effect on the computed velocity would it have to 
take 365.25 instead of 365 ? How would you correct your 
division for 365.25 as divisor after having obtained the quo- 
tient, dividing by 365? what change would using 365.26 in- 
stead of 365.25 effect in the computed velocity? 

5. The distance of the moon from the earth varies between 
221,000 and 260,000 miles, mean 238,000 ; discuss the length 
of the path of the moon and the velocity of the moon which 
has a periodic time of 27.32 days. 

6. A man whose salary is $ 3000 pays $ 480 for rent. 
What per cent is this of his salary ? Suppose that he earns 
S 275 in addition to his salary, what per cent is the rent paid 
of his income ? Compute only to tenths of one per cent. 

7. If a man with an income $ 3275 pays $ 1100 per annum 
for food, S 630 for clothes, $ 240 for life insurance, $ 200 for 
" higher life," and saves the balance, compute his budget by 
per cents. 

8. Given that a pendulum of length I cm. makes one beat, 
one oscillation, in t seconds, connected by the relation, 

find the length I to two decimal places when t = 1, 



36 UNIFIED MATHEMATICS 

9. What effect on I does a change from 980 to 981 pro- 
duce ? What decimal place in I would be affected ? • 

10. What error would the use of 3^ instead of 3.14159 
introduce ? 

11. The number n of vibrations of a pendulum of length 
99.39 cm. is 86400, when g = 980.96 ; g is the acceleration 
due to gravity, and the formula for the number of vibra- 
tions is given by the formula, 

86400 jg- 

or for the seconds' pendulum, when I = 99.39, g = 980.96, it 
is n = 86,400. Suppose that at the top of a mountain (g di- 
minishes) this pendulum of length 99.39 loses 86 beats per 
day, what is the approximate percentage of loss inn? The 
percentage of loss in g is approximately double this since 

Vl — 1 = 1 . What . is the approximate loss in g ? 

2 8 

Take 86 as y L of 1 % of 86,400. 

12. Given g == 980, I = 50, compute n in the formula of 
problem 11. What maximum effect on n would a change 
from I = 50 to I = 50.5 cm. produce ? 

13. Compute the weight of a table top, hardwood, dimen- 
sions correct to .05 foot, top 48.1 x 36.4 x 2.1 inches. Weight 
of wood 48 pounds per cubic foot. 

14. If a table top similar to the above weighs 97 pounds, 
compute the weight per cubic foot of the wood. 

15. If water weighs 62.4 pounds per cubic foot, Compute 
the specific gravity of each of the preceding woods. 

_ wt. of cubic foot of wood 
~~ wt. of cubic foot of water ' 

16. The path of the earth is approximately a circle of radius 
92.9 x 10 6 miles, of which the center is approximately 1-j- mil- 
lion miles from the sun. Compute this circumference and 
compare with the results in problems 1 and 2. 



APPLICATIONS OF ALGEBRA TO ARITHMETIC 37 



17. Factor the following, using wherever possible the factor 
theorem to determine factors : 

a. x* - 7a£ + 10 a,\ d. x 4 -Sx 1 - 20. 

6. 3 y 9 - - 5y + 2. e. v* - 7 v z + 8. 

c. 3?/ 2 -12. / xy 4 + x 4 y. 

18. Give the approximate square root of the following : 

a. 36.6. /. 98. 

b. 35.4. ^ 1.04. 

c. 104. fc. 1.12. 

d. 96. i 4.08. 

e. 126. j. 9.12. 

19. Complete the following : 



(x-2y)( 
(x-2y)( 
(x-2y)( 
(x-2y)( 
(x-2y){ 
{x-2y){ 



= a 2 - 4 2/2. 

= x ! — ixy + 4 j/ 2 . 

= j?-8}>. 

= x 2 — 6xy + 8yK 

= a 2 + 10 xy — 24 ?/ 2 . 

= 3x 2 -10*!/+8!/ 2 . 



CHAPTER III 

EXPONENTS AND LOGARITHMS 

1. Exponent laws. — For convenience the product of a by 
itself, a x a, is represented by a 2 , a x a x a by a 3 , ..., and a • a 
• a • a to m factors by a ra . In this notation m is called the ex- 
ponent and a the base. The following laws evidently hold : 

I. a m • a n = a m+n . 

II. — = a m ~ n , when m > w. 
a n 

III. (a OT ) n = a OTn . 
IY. (a • 6) m = a m • b m . 

In the definition as given, m represents the number of 
factors and is assumed to be a positive integer. However, it 
is found possible to define a m for all real values (fractional, 
negative, zero, irrational) of m so as to have the resulting 
numbers combine according to the four laws given above. 

Thus, a • a m = a 0+OT = a m , if Law I is to continue to hold ; 
hence, a must be defined to equal 1, since multiplying a 
number, a m , it gives that number. To be justified in using a 
zero exponent with this meaning the other exponent laws must 
be shown to hold when either m or n is zero, but in II only 
n could be zero at this point. 

For a negative integer, — n, if Law I is to hold, ar n must be 
defined as such a number that a n • a~ n = ar n+n = a° = 1 ; hence 

we define a~ n as the reciprocal of a n , a~ n = — • All the laws 

a n 

I to IV can be shown to hold under this extension of the 

meaning of a n . 

38 



EXPONENTS AND LOGARITHMS 39 

p 
Similarly, a q , if Law III is to hold, must represent a num- 

p 
ber which raised to the qih power equals a p ; a q is thus 

defined as the qth root of the jjth power of a. Taking this defi- 

p 
nition of a q , Laws I to IV can be shown to hold with this 

extension in possible values of m and n ; p and q are assumed 

to be integers. 

p mp 

For fractional exponents a fifth law is introduced, a 9 = a mq . 

For irrational values of n, a n is defined by a limiting pro- 
cess. Thus, a 1 is defined as the limit of the series a lA , a 1-41 , 
a 1414 •••, wherein the successive rational exponents define the 
square root of 2. 

The operations of elementary algebra with radicals are 
made subject to the exponent laws. 

Thus, 2 V3 = 2 • 3* = (2*)* • 3* = (2 2 • 3)* = 12* = V12, 
by successive application of V, III, and IV. 

The operation of raising to a power indicated by a m , with 
m integral, is called involution. The inverse operation of 
finding x when x m is given equal to a is called evolution. 

PROBLEMS 

1. Write as ordinary numbers, without exponents : 

10 2 , lO- 2 , 100*, 100*, 100-*, 10007, ioo. 

2. Find the approximate numerical value : 

10*, 10t, 10^ iQ-i io-f 5*, 2*, 5*, 5" 3 . 

3. Write the following expressions in the form, 10 B : 

Vio, Vio, * (JLY, i, i J-. 

Vio \Vip/ 10 105 

4. Which of the exponent laws are applied in simplifying 
the following expressions : 

V200, V2000, — ? 
V3 



40 UNIFIED MATHEMATICS 

5. What exponent must be applied to 10, as a base, to give 
1000 ? to give .001 ? to give 1 ? 

6. Simplify the following expressions : 

(10 3 ) 5 , (10 3 )(10 5 ); ij?, 2 3 -53; VWK 

2. Logarithms. — A logarithm is an exponent. 

The relation x = a m may be written m — log a x. 
m is the exponent which applied to a gives x ; 
m is the logarithm of x to the base a. 

3. Fundamental laws of logarithms. 

a. Logarithm of a product. 

If x =' a m and ?/ = a", 
x ■ y = a m • a n = a m+n . 
l°g a (a-?/)=m + ii = log a # + log a #. 

In the language of logarithms and translated into ordinary 
language this theorem is as follows : 

I. log a (*-y)=log a * + log a z/; 

in words, the logarithm of a product is the sum of the logarithms 
of the factors. 

b. Logarithm of a quotient. 

x a 171 x 

- = — = a m ~ n , log tt - = m — n = log a x — log a y. 

y a n y 

II. log a -=log a x-log a y; 

y 

in words, the logarithm of a quotient is the logarithm of the 
dividend minus the logarithm of the divisor. 

c. Logarithm of a power. 

If x = a m , x n = (a m ) n = a mn ; log a x n = m • n = n log a x. 
III. log a JT = n log a x ; 
the logarithm of a power of a number is the index of the power 
times the logarithm of the number. 



EXPONENTS AND LOGARITHMS 41 

Since our exponent laws hold for all values of m and n, these 
theorems hold for all values of m and n. 

x and y are assumed to be positive numbers and for compu- 
tation purposes 10 is commonly taken as the base. 

We assume that as the logarithm increases the number 
increases. This can be readily proved from the fact that 
10'" • 10" = 10"" +n ; no matter how small the n is, as a positive 
quantity, 10" is greater than 1. For n any positive fraction, 

■^ , 10 n represents the <?th root of the pth power of 10, wherein 
Q 

p and q are integers. Now 10 p will be an integer 10, or greater 
than 10, and the qth root of this integer will be a number 
greater than 1 as it will be a number which raised to the gth 
power equals 10, 100, 1000, or some greater number. It could 
not be less than 1, as every positive integral power of a number 
less than 1 is also less than 1. 

4. Logarithms, characteristic and mantissa. — Any two-place 
number lies between 10 and 100; the logarithm will lie be- 
tween 1 and 2. Any four-place number lies between 1000 
and 10,000 ; the logarithm will lie between 3 and 4, since as 
the number increases the logarithm increases. The fraction 
.07 is greater than .01 and less than .1 ; the logarithm is then 
greater than — 2 and less than — 1. 

The logarithm of any number between 1 and 10 is a fraction, 
expressed commonly as a decimal between and 1. 

The logarithm of any given number which is expressed in 
decimal notation can be expressed as an integer, positive or 
negative, called the characteristic, plus the positive decimal 
fraction, the mantissa, which is the logarithm of that number 
between 1 and 10, having the same succession of digits as the 
given number. Initial zeros are not included in the succession. 

Let k represent any number between 1 and 10, written in 
our ordinary decimal notation ; then 10" k, n any positive or 
negative integer, can represent any number written in decimal 
notation. 



42 



UNIFIED MATHEMATICS 



Exponential 




Column of 


Column of 


Column of 


Form 




Numbers 


Numbers : 


LOG &RITHM8 


10-1 2 


= 


.000000000001 


log .000000000001 


= -12 


10-5 


= 


.00001 


log .00001 


= - 5 


10-4 


= 


.0001 


log .0001 


= -4 


10-3 


= 


.001 


log .001 


= -3 


10-2 


= 


.01 


log .01 


= -2 


io-i 


= 




log.l 




100 


= 


1 


log 1 


= 


10i 


= 


10 


log 10 


= 1 


10 2 


= 


100 


log 100 


= 2 


103 


= 


1000 


log 1000 


= 3 


10 4 


= 


10000 


log 10000 


= 4 


10 5 


= 


100000 


log 100000 


= 5 


1012 


= 


1000000000000 


log 1000000000000 


= 12 





These positive numbers, middle columns, are arranged 
vertically in order of magnitude; the exponents (left) or 
logarithms (right) also are arranged vertically in order, in- 
creasing from — 12 (or from — oo indicated by dots above) to 
— 5, to — 4, to — 3, ••• to 0, to 1, ••• to 12, — to oo. As the num- 
ber increases the logarithm increases. Placing any number, 
not an integral power of 10, in its proper place as to magnitude 
on such a diagram, the logarithm has for integral part the 
logarithm of the preceding number in the table. Thus, 75.64 
falls between 10 and 100 and its logarithm will be 1 + ; .07564 
falls between .01 and .1 and its logarithm will be — 2 + , mean- 
ing — 2 plus some positive fraction. 



EXPONENTS AND LOGARITHMS 43 

log 10" k = log 10" + log k = n log 10 + log k = n -f log k. 
n is the characteristic ; log k is the mantissa. 

Thus, log 3.16229 = .50000, since 3.16229 is the approximate 
square root of 10. 

log 31622.9 = log (10) 4 (3.16229) = log 10 4 + log 3.16229 
= 4 + log 3.16229 
= 4.50000. 
log .000316229 = log (10)" 4 (3.16229) = log (10)~ 4 + log 3.16229 
= _ 4 + .50000 

= 4.50000, only the 4 is negative 
= 6.50000 - 10, since - 4 equals 6-10 
= 16.5000 - 20 
= 26.50000 - 30, 



The minus sign over the 4 indicates that only the charac- 
teristic is negative ; the alternate forms for writing a negative 
characteristic are frequently found convenient to use, particu- 
larly in extracting roots and with the trigonometric functions. 

Kule. — TJie logarithm of any number greater than unity has 
as characteristic a positive integer (0 included) which is 1 less 
than the number of digits to the lejt of the decimal point. 

TJie logarithm of any decimal fraction has as characteristic 
the negative of a positive integer (0 not included) which is 1 
greater than the number of zeros between the decimal point and 
the first significant digit (i.e. digit other than 0) to the right of 
the decimal point. 

5. Computation of logarithms. — Logarithms are actually cal- 
culated by a series derived from formulas obtained in higher 
mathematics. However, a simple although laborious method 
of computing logarithms approximately may make the signifi- 
cance of the logarithm somewhat clearer. 



44 UNIFIED MATHEMATICS 

2 10 = 1024, 2° = 1,048,576. 

Evidently, 10 6 < 2 20 < 10 7 , since 1,048,576 is greater than 
1,000,000 and less than 10,000,000. 
Extracting the twentieth root, 

10 6 < 2 20 < 10 7 

gives 10^ < 2 < 10 & 

or 10- 30 < 2 < 10 35 . 

Hence, 2 is greater than 10 with an exponent .30, and less 
than 10 with an exponent .35. 

2 40 = (1,048,576) 2 < 1,100,000,000,000 
> 1,000,000,000,000 ; 

2 80 < 1.21 x 10 24 (2 40 by 2 40 ) and greater than 1 x 10 24 . 

Whence 2 100 < 1.33 x 10 30 and greater than 1 x 10 30 , whence 
10 30 < 2 100 < iO 31 , whence 

10- 30 < 2 < 10* or log 10 2 = .30+. 

Similarly, 3 20 is 3,486,784,401, or greater than 10 9 and less 
than 10 10 ; hence, log 3 lies between .45 and .50, the computa- 
tion of 3 20 is easily made by using 3 4 = 81 and 3 20 = (81)5, mu i_ 
tiplying each time by 81 instead of by 3 ; the partial product 
by the 1 does not need to be rewritten. 

6. Tables of logarithms. — The exponents or logarithms to 
the base 10 of all numbers up to 100,000 have been computed 
by methods of higher mathematics ; these logarithms are 
arranged in tabular form in the natural order of the corre- 
sponding numbers, so as to be convenient for computation 
purposes. Our tables give the mantissas of the logarithms of 
all numbers between 100 and 999 ; by the insertion of the 
proper characteristic the logarithms of all numbers having 
one, two, or three significant figures, i.e. disregarding initial 
and terminal zeros, are given by our tables. 



EXPONENTS AND LOGARITHMS 45 

EXERCISES ON TABLE OF LOGARITHMS, PP. 498-499 

1. Find the logarithms of the following numbers, writing 
first the proper characteristic, following the rules given on 
page 45, before employing the table. 

a. log 234 = e. log 6.06 = i. log .00032 = 

b. log 46900= f. log 1000 = j. log .999 = 

c. log 2.91 = g. log .543 = k. log .001 = 

d. log 8450 = h. log .00902= I. log 3 = 

2. Find numbers corresponding to the following logarithms : 

a. log =0.0414 g. log =9.7396-10 

b. log = 2.3096 h. log = 6.9996 - 10 

c. log =4.8500 i. log =8.9031-10 

d. log =6.6425 j. log =5.8904-10 

e. log =3.0000 k. log =9.0086-10 
/. log =1.3010 I. log =7.3010-10 

3. G-iven log 2 = .3010, write the equivalent statement in 
exponential form. 

4. Given 10 699 = 5, write the equivalent statement in terms 
of logarithms. 

5. Given 10- 699 = 5, and 10* = 2, multiply and find the value 
of x. 

7. Logarithms. Interpolation. — The process employed in 
extending the use of a table of logarithms of numbers with 
three significant figures, so as to give logarithms of numbers 
having four significant figures, is called interpolation. The 
method applies to increase in a similar manner the scope of 
any table of logarithms so as to give the logarithms of num- 
bers having at least one more significant figure beyond those 
of the numbers in the tables. 

The numerical process employed in interpolation may be 
illustrated graphically. 



46 



UNIFIED MATHEMATICS 



204 



Given a trapezoid with bases a and 6, and altitude h, we wish to divide 
the altitude into 10 equal parts and to find the lengths of the dividing 

lines parallel to the bases. 
Evidently each of these 
lines differs from the pre- 
ceding line by one-tenth 
of the difference, b — a 
= d, between the bases. 
The upper line b may 
represent the logarithm 
of some number, e.g. of 
204 ; the lower line may 
represent the logarithm 
of a number smaller by 
one unit, e.g. of 203 ; 
d = b — a represents 
then the difference of the 
logarithms. We assume 
then that the nine inter- 
vening lines represent 
approximately the loga- 
rithms of 203.1, 203.2, 
... up to 203.9. 



h 




]</ = & — a/ 


! / 


i / 


i / 


; / 


1 / 


i / 


• / 


_i/ 


y 


U A 



203 



Graphical Interpolation between 75 and 96 (2.3075 
and 2.3096), nine values interpolated 



N 





1 


2 


3 


4 


5 


6 


, ? 


8 


20 


3010 


3032 


3054 


3075 


3096 


3118 


3139 


3160 


3181 


21 


3222 


3243 


3263 


3284 


3304 


3324 


3345 


3365 


3385 



9 

3201 
3404 



This section of a logarithm table gives the logarithms of all 
numbers with three significant figures, having the succession 
of digits beginning 20 or 21. Note that the differences begin 
at 22 in the first line and drop to 19 in the second. 



log 203 
log 203.1 
log 203.2 

to 
log 203.9 

log 204 = 2.3096 



2.3075 

1 

i 
1 



These logarithms lie between 2.3075 and 
2.3096 ; they increase steadily ; we assume that 
they increase by uniform amounts, which we see 
in the table is roughly true for numbers from 200 
to 219. The uniform increase of these logarithms 
of 203.1 to 203.9 is ^ of the difference between 
the logarithms of 203 and 204, i.e. 2.1 of the 
units in the last place of our logarithms. It 



EXPONENTS AND LOGARITHMS 



47 



would not be correct to increase the number of places in our logarithms 
as our process is only an approximation not correct to further places even 
as our logarithms are only approximations to four decimal places. The 
logarithm of two is to five places .30103, to ten places .3010299957, to 
twenty places .30102999566398119521. 

log 203.0 = 2.3075 

203.1 = 2.3077 

203.2 = 2.3079 



203.3 = 2.3081 
203.1 = 2.3083 

203.5 = 2.3086 

203.6 = 2.3088 

203.7 = 2.3090 

203.8 = 2.3092 

203.9 = 2.3091 
201.0 = 2.3096 



log .203 =1.3075 
.2031 = 1.3077 

.2032 = 1.3079 
.2033 = 1.3081 
.2031 = 1.3083 
.2035 = 1.3086 
.2036 = 1.3088 
.2037 = 1.3090 
.2038 = 1.3092 
.2039 = 1.3094 
.2010 = 1.3096 



A four-place table of numbers, with logarithms to five 
places, gives simply : 





j I 


2 


3 


4 


5 


6 


7 


8 


9 


203 
204 


30750 30771 
30963 j 30984 


30792 
31006 


30814 
31027 


30835 
31048 


30856 
31069 


30878 
31091 


30899 
31112 


30920 
31133 


30942 
31154 



The appropriate characteristic must be inserted by the com- 
puter. A careful examination will show that our interpolation 
gives an incorrect four-place result for 203.4, with an error of 
half a unit. This type of error is inevitable, using four-place 
tables. In general, such errors tend to equalize each other ; 
where absolute accuracy to four places is necessary, five- or 
even six-place tables must be used. Even with four-place tables 
it is evident that with a difference, called the tabular difference, 
of 21 to 21 the normal difference will be 2 units and 1 to 4 
extra units will have to be distributed in the addition. 

In the illustration above it may be noted that log 203.4 as 
2.30835 does not inform us, strictly, as to whether log 203.4 to 
four places is 2.3083 or 2.3084; the latter is the case here 



48 UNIFIED MATHEMATICS 

since if log 203.4 is found to further places than live the fifth 
decimal place is actually a 5. Whenever the five-place loga- 
rithm is a terminal 5 which has been obtained by increasing 
a 4 to 5 the four-place logarithm would not be increased by 
one unit ; thus, log 2007 to 5 places is 3.30255, to 6 places 
log 2007 is 3.302547, and hence to four places log 2007 is 
3.3025. In some tables of logarithms a terminal 5 due to an 
increase, as in log 2007 = 3.30255, is marked with a super- 
imposed negative sign or other mark. 

In physical problems, measurement to three places permits 
of computation to three places only ; the fourth place by inter- 
polation assures accuracy, in general, of the third place. 

The reverse process is to find the number, given the loga- 
rithm ; thus, to find the number whose logarithm is 2.3088, we 

see that 

Diff. 21 

the table gives log 203 = 2.3075 1 1 2.1 

and log 204 = 2.3096 2 4.2 

The tabular difference is 21 3 6.3 

The given logarithm is 2.3088 4 8.4 

The difference between it and the 5 io.5 

smaller logarithm nearest to it is 13 6 12.6 

7 14.7 

8 16.8 

9 18.9 

The table of tenths of the tabular difference, which is fre- 
quently given in tables of logarithms, shows that 13 is nearest 
to .6 x 21. The number is 203.6. Had the given logarithm 
been 2.3087, we would find as the number again 203.6, since 
the actual difference is 12, which also is nearer to .6 of 21 than 
to .5 of 21. Note that in the table, and in the difference 
table, the logarithm or part of the logarithm lies always to the 
right ; the number, or part of the number, lies to the left or 
above. 

8. Historical note. — Fundamentally the notion of logarithms 
is intimately connected, as we have shown, with the notion of 



EXPONENTS AND LOGARITHMS 



49 



exponents. The one-to-one correspondence between exponents 
and a series of successive powers of a given number was noted 



Exponents 


-4 


-3 


-2 


- 1 




1 


1 

2 


2 
4 


3 

8 


4 


5 


6 


7 


8 


9 


10 


Numbers 


16 


i 

8 


I 

4 


1 
2 


16 


32 


64 


128 


256 


512 


1024 



Figure from Stifel's Arithmetica Integra, 1544 



many years before logarithms were developed, in many arith- 
metical textbooks of the fifteenth and sixteenth centuries. 
A German mathematician, Stifel (1486-1567), published a 
work entitled Arithmetica Integra, 1544, in which these " ex- 
ponentes," as he termed them, were extended to the left. 
Stifel spoke of the great possible use of such series for compu- 
tation in which addition would replace multiplication, and sub- 
traction replace division ; but he developed the idea no further. 

In 1614 John Napier, Baron of Merchiston, a Scotchman, 
revolutionized computation processes by the composition of 
logarithmic tables, based on the idea of the comparison of two 
series essentially of the kind indicated above. The adoption 
of 10 as the base of a logarithmic series was due to a friend of 
Napier, Henry Briggs, who published in 1617 the decimal 
logarithms of the first thousand numbers. 

In recent years the widespread adoption of computing 
machines which carry multiplications and divisions to fifteen 
and twenty places is somewhat replacing logarithms in the 
offices of great insurance companies and, to some extent, in 
observatories. 

Logarithms. Illustrative Problems. — I. Find by four- 
place logarithms : 

, 2.03 137 



a. 203 x 137 ; 

a. log 203 = 2.3075 

log 137 = 2.1367 

log product = 4.4442 

product = 27810 



13. 



c. 



2.03 



d. (203) 3 . 



By interpolation since 2, the difference 
found, is in tenths nearest to .1 of 16, the tab- 
ular difference. 



50 



UNIFIED MATHEMATICS 



b. log 2.03 

log 137 

log quotient 



0.3075 
2.1367 
2.1708 



log 2.03 
log 137 
log quotient 



10.3075 - 10 
2.1367 

10 



8.1708 
quotient = .01482 
The 2 is obtained by interpolation since 
5, the difference found, is nearest to .2 of 
29, the tabular difference. 



c. log 137 =2.1367 
log 2.03= .3075 
log quotient = 1.8292 
Quotient is 67.48, since the differ- 
ence 5 is in tenths nearest to .8 of 6, 
the tabular difference. 



d. 



log 203 = 2.3075 
3 



log (203)3 = 6.9225 
(203)3 = 8,366,000 



If these numbers represent physical constants, obtained by 
measurement, our computations would be slightly more ac- 
curate than the measurements warrant. Even in prices, in 
general, these results would be sufficiently accurate for pur- 
poses of business. Thus 137 pounds of cheese at 20.3 ^ per pound 
would be found to be worth $ 27.81 ; 137 tons of mine-run 
coal at $ 2.03 per ton is worth $ 278.10, with a possible error 
of 10^; 1370 tons of coal at $2.03 would be $2781 with a, 
possible error of 50^, which is negligible in business of this 
magnitude. 

II. Find by logarithms, four-place, to four places : 



a. 203.8 x 137.5 ; b. 

a. log 203. 8 = 2.3092 
+ loff 137.5 = 2.1383 



13750 



log product = 4.4470 
product = 27.790 

c. log 203.8 = 2.3092 

3 

log (203.8)3 = 6.9276 
(203.8)3 = 8,464,000 



; c. (203.8)3; d, (.02038)^ 

6. log 20.38 = 1.3092 
- log 13750 = 4.1383 
log quotient = 3.1709 
or 7.1709 -10 
quotient = .001482 
By interpolation the given differ- 
ence of 6(.1709 — .1703) is nearest 
in tenths to .2 of the tabular dif- 
ference, 29 (.1732 - .1703). 



EXPONENTS AND LOGARITHMS 51 

d. log (.02038)* = | log .02038 

log .02038 = 2.3092 8.3092 - 10 

3 ° r 8 

5 6.9276 24.9276-30 

The division by 5 causes trouble because of the negative characteristic. 
To avoid the difficulty, write 6.9276 as 44.8276 — 50. Dividing this by 5 
gives 8.9855 - 10 or 2.9855. 

(.02038)* = .09672. 

EXERCISES 

1. Using a watch with a second hand, time yourself on 
looking up the logarithms of the following 20 numbers ; as a 
preliminary exercise look up the logarithms of these numbers 
without using the fourth significant figures, not requiring 
interpolating. 



log 


314.6 = 


log 


813.2 = 


log 


5.462 = 


log. 


003468 = 


log 


.3085 = 


log 


769.9 = 


log 


67870 = 


log 


368.60 = 


log 


53.85 = 


log 


19.26 = 



log 


14.32 


log 1876000 


log 


81930 


log 


.08764 


log 


3.250 


log 


.7263 


log 


200.4 


log 


399.8 


log 


210.4 


log 


.03899 



Twenty minutes should be the maximum time on this list ; practice 
until you can do all 20 with interpolations within 15 minutes or even 10 
minutes ; all 20 characteristics should be written before you begin to use 
the tables. 

2. Find by four-place tables the logarithms of the following 
numbers, interpolating: 326, .08342, 10,050, .008766, 5499, 
3.482 x 10*, 37.04, 290.40, .9647, 38.55, .06948, 3001, 2.777 x 
10- 6 , 784.4, 6,934,000, 5.341, 70.98, .1237, 8462, 3740. 

Time yourself ; the 20 should not take more than 12 minutes. 



52 UNIFIED MATHEMATICS 

3. Find the numbers corresponding to the following loga- 
rithms — no interpolation is necessary. Take the time of 
looking up the numbers and writing these in a prepared form 
as answers ; the time should not exceed 8 minutes. 



log 


= 3.8414 




log 


= 1.0492 


log 


= 0.9996 




log 


= 5.9063 


log 


= 2.6866 




log 


= 2.9557 


log 


= 1.7482 




log 


= 3.1732 


log 


= 6.3010 




log 


= 6.2672 


log 


= 8.0086 - 


-10 


log 


= 5.7832 


log 


= 4.2856 




log 


= 1.1761 


log 


= 1.8837 




log 


= 9.5024 


log 


= 9.3201 - 


-10 


log 


= 2.4786 


log 


= 2.7789 




log 


= 3.1673 



10 



The time includes only looking up the logarithm and writing it down; 
it should not include copying the problem. A piece of blank paper may 
be placed alongside of each column and the logarithms written upon the 
paper ; by folding the paper lengthwise the two columns may be placed 
upon the same sheet. Practice with timing. 

4. Find the numbers corresponding to the following, loga- 
rithms, interpolating wherever necessary ; time yourself, ir 
looking up numbers, first writing the given logs in column 
form, and not counting that time. The 20 should not take 
more than 15 minutes ; 10 minutes is slightly better than 
average time. 



log 


= 3.5861 


log 


= 8.5418 - 


-10 


log 


= 5.6427 


log 


= 2.0923 




log 


= 1.4436 


log 


= 2.9844 




log 


= 4.7320 


log 


= 3.3080 




log 


= 6.9428 


log 


= 1.218 




log 


= 2.4415 


log 


= 7.8419 - 


-10 


log 


= 6.4893 - 10 


log 


= 0.4630 




log 


= 5.8662 


log 


= 1.7848 




log 


= 1.5729 


log 


= 0.9618 




log 


= 2.9990 


log 


= 1.7276 





EXPONENTS AND LOGARITHMS 53 

5. By chaining, the sides of a rectangular field are found to 
measure 413.2 feet and 618.4 feet. Find the area in square feet 
and in acres. What effect upon the computed area does an 
error .1 of a foot in measurement make ? Consider this 
fact in making the computation, not assuming that these 
lengths are more accurately given than to .1 of one foot. 



2.7455 


4.3925 


0.0420 


6.4105 


1.2222 


2.9213 


1.3660 


2.0621 



6. Use Hero's formula, A = Vs(s — a) (s — b) (s — c), 

wherein a, b, c are the sides and * — — — ^t_w = s , to compute 

Z 

the area of a triangle whose sides are 413.2 feet, 618.4 feet, 
and 753.2 feet. Discuss errors as before. Do all the prelimi- 
nary computation of s, s — a, s — b, s — c before looking up 
any logarithms. 

7. Time yourself on finding the 20 numbers corresponding 
to the following logarithms : 

2.1120 1.2480 1.9462 

1.6150 5.4151 0.5132 

8.9312 - 10 3.5674 3.3808 

4.8990 1.1174 7.8973 - 10 

8. Perform the following computations, using logarithms : 
a. 54.04 x 376.2; b. ||g; c. (54.04)^(3.762); 

d. V5T04; e. V54.04 x 376.2. 

9. Find the volume in gallons of a cylindrical can, 18 inches 
in diameter and 30 inches high. (1 gallon = 231 cubic inches.) 

10. Show that the capacity of a cylindrical can in gallons 
can be written as i of 1 % of dVi + 2 % of the i of 1 % of dVi, 
or as 1.02 x 0.00J x d 2 h, given d and h in inches. 

11. Find the volume in cubic feet of a silo 16 feet in diame- 
ter and 32 feet high. Compute for 15.9, 15.95, 16.05, and 16.1 
feet in diameter. If the measurements are correctly given 
within .1 of one foot, how accurately can the volume be given, 
with a 16-foot diameter ? 



54 UNIFIED MATHEMATICS 

12. Find approximate value of 2 50 , 2 -50 , 2^, and 2~ T ^ by 
logarithms. 

13. Extract the cube roots of 2, 3, 4, 5, 6, 7, and 8 by 
logarithms. Multiply the value of 2i by the value of 3^, and 
compare with your value of 6i. 

14. Given a pendulum of length Z centimeters and time of 
oscillation t seconds, you have the following formula connect- 
ing I, and t: \—j~ 

Given t = l, compute 2; given 1 = 60, compute t; given that 
t = 1 but that instead of 980 you have 981, compute I. 

15-30. Solve the problems at the end of Chapter II using 
logarithms. 

9. Compound interest. — When interest is added to the prin- 
cipal at the end of stated intervals forming a new principal 
which is to continue to draw interest, the total increase in 
the original principal which accumulates by this process con- 
tinued for two or more intervals is called the compound inter- 
est on the original principal. 

Let P represent the principal, i the interest rate per inter- 
val, and n the number of intervals, and S the amount at the 
end of n intervals. Given interest compounded at rate i per 
annum for n years. 

At the end of 1 year you have P + iP = P(l -+- i). 

At the end of 2 years you have 

P(i + i) + iP(i + o = P(i + 2 - 

At the end of 3 years you have 

P(l + if + iP(l + 2 = p (! + *) 3 - 



At the end of n years you have S = P(l -f- *) n - 



EXPONENTS AND LOGARITHMS 55 

Or you may say that since the interest for 1 year in- 
creases the principal P to (1 + i)P, then in 1 further year this 
new principal P(l -f- i) will be increased in the same ratio, 
giving P(l +i) X (1 + i) or P(l + f) 2 , and for each further year 
the factor (1 + i) is introduced. Hence the amount at the end 
of n years is P(l + /)". 

If interest is compounded at the end of every three months, 

or every six months, you substitute for i, - or -, and for n, 

4 n or 2 n, since the number of intervals of three months in n 
years is 4 n and of six months is 2 n. 

The formula S = PI 1 + — ] is used for an interest rate 

V w 

given as j per annum, but compounded m times per annum, 

at rate ^- for each interval. 
m 

Problems in compound interest lend themselves to solution 
by logarithms. 

Given, S = P(l+iy. 

log£ = logP+log(l + i;r. 
log S = log P + n log (1 -f- i). 
log P = log ^ — n log (1 -f i). 

^^ log /S^-logP 

log (1 + i) 

log(l + = lQ g^- lo g P . 

Note that it is better not to use these as formulas, memoriz- 
ing them, but rather to go back to the fundamental relation, 

5 = P(l + i) n . Note also that the formula holds for other 
than integral values of n ; thus at 6 % per annum the interest 
on the amount P for six months or eight months is defined as 
P(l + .06)* - P or P(l + .06)1- P, respectively. Hence for 
«t-f i years the amount would be P(l + i) n (l + i)% =P(1-M) W+ ^ 



56 UNIFIED MATHEMATICS 

PROBLEMS 

1. Find the amount of $ 1000 at interest 4 % annually, 
compounded for 20 years. Find the amount when compounded 
semi-annually at a nominal rate of 4 % per annum, i.e. 2 °J 
semi-annually. 

2. In how many years will money double itself at 4 %, 
5 % , 6 °f interest, compounded annually ? 

3. Given that at the end of 20 years $ 1000 amounts to 
$ 1480, what is its approximate rate of interest ? 

4. Given that at 5 % interest, compounded annually, $ 1000 
amounts to $ 1480, what is the approximate number of years ? 

5. Find the compound amount of $ 1400 at 5 % interest, 
compounded semi-annually, for 10 years, 11 years, 12 years, 
up to 20 years. 

6. If $ 100 is left to accumulate at 3 % interest, com- 
pounded annually, what will it amount to in 100 years ? Solve 
by logarithms. What amount put at 3 % interest will amount 
in 100 years to $ 1,000,000 ? What is the present equivalent 
of $ 1000 to be paid at the end of 100 years, money worth 3 % , 
compounded annually ? 

7. Solve problem 6 for 4%, 5%, and 6% interest, com- 
pounded annually. For 6 % per year, compounded semi- 
annually, for 50 years. 

8. Benjamin Franklin, who died in 1790, left 1000 pounds 
to " the town of Boston " and the same to the city of Phila- 
delphia. His will directed that this amount should be loaned 
at interest to young artisans, and thus accumulated for 100 
years until the principal should have increased to 130,000 
pounds. He directed further that at that time the major por- 
tion of this amount should be expended for some public im- 
provement and the residue left to accumulate, similarly, for 
another hundred years. What rate of interest did Franklin 
assume that his money would earn ? In Boston the amount, 



EXPONENTS AND LOGARITHMS 57 

$ 5000 approximately, accumulated to about $ 400,000. Find 
the average rate of interest earned annually. Assuming that 
S 5000 was kept aside in 1891, as directed, what will this 
amount to in 1991, compounded at 4 % annually ? 

9. Find the amount at the end of 200 years of $ 5000, 
interest at 4 %, 5 %, and 6 %, compounded annually. 

10. If a business doubles its capital, out of earnings, in 
12 years, what rate of interest on capital invested does this 
represent per year ? If in 20 years the capital is doubled, find 
the rate of interest earned. 

11. The United States has increased in population from 7.2 
million in 1810 to 101.1 million in 1910 ; find the approximate 
rate of increase per year, and for each ten-year period. Com- 
pare with the figures on page 65. 

12. The city of New York increased in population from 
120,000 to 4,769,000 in the interval from 1810 to 1910. Com- 
pute the average annual rate of increase, using the formula, 
120,000(1 + 100 = 4,769,000. Compute the average ten-year 
increase and compare with the actual statistics on page 66. 



CHAPTER IV 
GRAPHICAL REPRESENTATION OF FUNCTIONS 

1. Functional relationships. Expressions of the form 3 # + 5, 
ax + b are called linear or first degree functions of the variable 
x ; in elementary algebra such expressions have been combined 
according to the fundamental operations and subject to the 
laws given in a preceding chapter. Further, some atten- 
tion is given in elementary algebra to expressions of the 
form ax 2 + bx -f c, the general quadratic function of x, and 
expressions involving higher powers of x. The expression 
ax n + bx 71 ' 1 -f- ••• is called an algebraic function of x of the 
nth degree when n is a positive integer and the coefficients 
a, b, ••• are constants. This represents of course a number for 
any value of x. 

F(x), G(x), <j>(x), A(x), ••• are methods of representing 
functional relationships ; F(x), (read F of x or F function of x), 
means that this expression assumes various values as x varies, 
these values being determined by some law. In the equation, 
y = 3" x -+- 5, y is explicitly given as a function of x ; y is here 
a linear function of x. In the equation, y = x 2 + 4 x — 5, y is 
an explicit function of x ; as # varies, so does y. In x 2 + y 2 = 25, 
as x takes on different values so does y, but one must solve for 
the corresponding values of y. Here y is called an implicit 
function of x. 

When two variable quantities are so related that the varia- 
tion of one of these depends upon the variation of the other, 
either is said to be a function of the other. Thus the pro- 
duction of wheat in the United States from 1900 to 1915 is a 
variable quantity depending upon the year of production. 
The height of a given tree is a function of its age; to each 
number expressing in any convenient unit of time the age of 

58 



GRAPHICAL REPRESENTATION OF FUNCTIONS 59 

the tree corresponds a given number expressing the height 
of the tree. Similarly the weight of a tree is a function of 
the age of the tree. This type of relationship cannot be ex- 
pressed algebraically. It may be exhibited by the two series 
of numbers, or it may be expressed graphically. 

2. Graphical representation of statistics. — Since two variable 
quantities are to be represented, two sets of numbers must be 
indicated ; this could be done by placing the two sets upon 
two lines straight or curved, drawn parallel to each other. 
This is the form used upon grocers' scales wherein the vari- 
ables of weight and corresponding price are placed upon con- 
centric circular arcs ; corresponding numbers are cut by the 
pointer. ^^^^^^^^^^^^^^_ 



Series of corresponding numbers graphically represented 

It is commonly more convenient to place the two scales for 
representing the two variable quantities upon two lines per- 
pendicular to each other. Upon the following figure the tem- 
perature and barometric pressure are indicated by the diagram 
for the week, March 4-11, 1918, at Ann Arbor, Michigan. 



60 



I'XIFIED MATHEMATICS 



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Temperature and barometric chare by moving pointer 
The sharp break in barometer curve corresponds to a violent rainstorm, 

The horizontal displacement of any point on either graph, located by 
the vertical rulings, indicates the time of the observation ; the correspond- 
ing temperature or pressure is indicated by the vertical displacement. 

ILLUSTRATIVE EXERCISES 
1. Production and price of wheat in the IT". S. from 1895 to 
1916 are given in statistical form and graphically. 



Production i Exports Pei- e 



Production Exports 



Pbice 



Teas 



Ykab 





Millions of Bushels 


Cents 




Millions of Bush - 




1895 


467 


126 


50.9 


1907 


034 


163 


B7.4 


1896 


428 


145 


72.':; 


1908 


665 


114 




1897 


530 


217 


80.8 


1909 


737 


B? 




1898 


675 


222 


58 2 


1910 


• 135 


69 


38 


1899 


547 


186 


58.4 


1911 


621 


80 


B7.4 


1900 


522 


216 


61.9 


1912 


730 


143 


' 


1901 


74S 


23o 


02.4 


1913 


763 


146 


79.9 


1902 


670 


203 


63.0 


1914 


891 


332 


98.6 


1903 


638 


121 


69.5 


1915 


1,026 


243 


91.9 


1904 


552 


44 


92.4 


1916 


640 




160.3 


1905 


693 


98 


74.8 


1917 








1906 


735 


148 


" 











Statistics from the Yearbook of the U.S. Department of Agriculture 



GRAPHICAL REPRESENTATION OF FUNCTIONS 61 





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Graphical representation of wheat production (continuous line) and price 
(broken line) in the United States, 1895-1917 

Note- that the graphical form of these statistics brings out several points 
of interest. In the first place the maximum price paid for wheat in the 
interval is immediately found, and so also the minimum price of 51 ^ 
(50.9^) in 1895. Further, the diagram shows very pointedly that a 
large production under normal circumstances is accompanied by a fall in 
price, and an immediate diminution of production. In 1917, under war 
conditions, both production and price increased greatly. 



2. The weight of water per cubic foot, or 60 pints, is 62.4 
pounds. For cylindrical vessels filled to a height of 12 inches 
the weight for an area 144 square inches in the base would be 
6//. 4 pounds ; for 72 square inches in the base the weight of 



62 



UNIFIED MATHEMATICS 



water would be 31.2 pounds ; for square inches the weight is 
pounds. On coordinate paper represent square inches of 
base on the horizontal line, taking 1 major division to repre- 
sent 10 square inches, and represent weight on the vertical 
line. 



Weight 

in lbs. 

100 




10 20 30 40 50 60 70 80 90 100 lip 120 130 140 
Square inches in base 

Weight of water in cylindrical vessels with varying base when filled to a 
height of 12 inches or 10 inches 

Note that the weight of 12 inches of water in a vessel with a base con- 
taining 50 square inches is 21.5 pounds, and conversely if a cylindrical 
vessel contains 21.5 pounds in 12 inches of height, the base contains 50 
square inches, and similarly, of course, for other values. 



PROBLEMS 

1. Plot the temperature, as vertical lengths, and the time, 
by hours, as horizontal lengths, for 24 hours. 

2. Plot the contents in pints of cylindrical vessels 12 
inches in height, with varying bases ; take that with base 144 
square inches, the capacity is 60 pints j with 72 square inches 



GRAPHICAL REPRESENTATION OF FUNCTIONS 63 

in the base, 30 pints ; with base, pints. The straight line 
joining these points can be used to give the base in square 
inches of any cylindrical vessel whose capacity for a height of 
12 inches is known. What would be the base of a vessel that 
contains 10 quarts when filled to the height of 12 inches? 

3. Plot cubic inches against pints, taking 1728 cubic inches 
as 60 pints. 

Increase in Volume with Temperature Increase 

As liquids are heated the volume changes, generally increasing ; thus 
water increases in volume when heated except between and + 4° C. 
Given 1000 cu. cm. of water at 4° C. and 1000 cu. cm. of mercury at 0° C, 
the volume at other temperatures is given by the following table : 



Temperature 


Volume of Water 


Volume of Mercury 


0° 


1000.13 


1000.00 


1° 


1000.06 


1000.2 


4° 


1000.00 


1000.9 


8° 


1000.13 


1001.4 


10° 


1000.27 


1001.8 


15° 


1000.87 


1002.7 


20° 


1001.77 


1003.6 


25° 


1002.94 


1004.5 


30° 


1004.35 


1005.4 


35° 


1005.98 


1006.3 


40° 


1007.82 


1007.2 



4. Plot the increase above 1000 cu. cm., or decrease, in cu. 
cm. in volume of the water, using 1 half -inch for 5° on hori- 
zontal axis and 1 half-inch for 1 cu. cm. on the vertical axis. 
Note that by adding 1000 to the given readings, actual volumes 
can be read. 

5. Plot the same curve for the increase in volume of the 
mercury. It is evident that the increases in volume of the 
mercury are approximately proportional to the increases in 
temperature. 



64 



UNIFIED MATHEMATICS 



Statistics ox Weight axd Height 

From an investigation of the statistics giving characteristics of a group 
of over 200,000 men and 130,000 women, the following facts are obtained 
on average height. The facts are given for groups of 1000. 





Frequency or No. in 
Group 




Men 


Women 


4' 9" 





1 


4' 10' 





4 


4' 11" 





10 


5' 0" 


2 


40 


5' 1" 


2 


55 


5' 2" 


5 


107 


5' 3" 


12 


135 


5 / 4// 


30 


184 


5' 5" 


55 


167 


5' 6" 


99 


134 


5' 7" 


127 


83 


5' 8" 


169 


48 


5' 9" 


145 


18 


5' 10" 


147 


8 


5' 11" 


104 


3 


6' 0" 


66 


1 


6' 1" 


22 





6' 2" 


11 





6' 3" 


3 





6' 4" 


1 






Weight (to Nearkst 

Integer in 5 or 0) of 

Mex ; Ages 35-39 ; 

Height l r 10" 


Frequency or 

Number in 

Group 


125 


4 


130 


14 


135 


33 


140 


60 


145 


78 


150 


114 


155 


95 


160 


106 


165 


90 


170 


87 


175 


72 


180 


59 


185 


48 


• 190 


37 


195 


25 


200 


32 


205 


12 


210 


14 


215 


4 


220 


8 


225 


3 


230 


1 



In any such group the number of individuals having any given char- 
acteristic is called the frequency corresponding to the given characteristic. 



6. Plot the frequency curve of heights of men and women, 
taking \ inch as corresponding to 1 inch of height on the 
horizontal axis and taking -i- inch vertical for 30 individuals. 
This curve represents very nearly what is termed a normal 
symmetrical distribution. 



GRAPHICAL REPRESENTATION OF FUNCTIONS 65 

7. Plot the frequency curve for weights of meu between 
35 and 39. 

Agricultural Statistics 

In the Yearbook of the Department of Agriculture statistics of produc- 
tion and prices of standard crops and farm products are given, covering a 
period frequently of 50 years. Use this Yearbook to obtain the data for 
the following curves : 

8. Plot the curve showing the production of corn in the 
United States from 1866 to the present time. Use 200,000,000 
bushels as a vertical unit, taking \ inch as the unit ; take one 
year as ^ of an inch. 

9. Plot prices on the diagram of 8, using a right-hand scale. 

10. Plot similarly statistics for the amount and price of 
sugar produced in the United States. 

11. Plot the average price in the United States of eggs by 
months for the current year ; plot butter prices similarly. 

Population Statistics 

The population statistics of the United States by 10-year intervals as 
given by the Statistical Atlas of the U. S. Bureau of Census are : 



Date 


U. S. 


New York 


Texas 


(Millions) 


(Thousands) 


(Thousands) 


1790 


3.9 


340 




1800 


5.3 


589 




1810 


7.2 


959 




1820 


9.6 


1,373 




1830 


12.9 


1,919 




1840 


17.1 


2,429 




1850 


23.2 


3,097 


213 


1860 


31.4 


3,881 


604 


1870 


38.6 


4,383 


819 


1880 


50.2 


5,083 


1,592 


1890 


63.0 


6,003 


2,236 


1900 


77.3 


7,263 


3,049 


11)10 


101.1 


9,114 


3,897 


1920 


117.0 


10,942 


4,734 



66 



UNIFIED MATHEMATICS 



12. Plot the curve of population of the United States. 

13. Plot the population curve for Michigan, and estimate 



the population for the 5-year periods. 



Date 


Michigan 
(Thousands) 


Date 


New Yobk City 
(Thousands) 


Date 


Michigan 
(Thousands) 


New York City 
(Thousands) 


1837 


175 


1790 


49 


1864 


804 








1800 


79 


1870 


1,184 


l v 478 






1810 


120 


1874 


1,334 








1820 


152 


1880 


1,637 


1,912 






1830 


242 


1884 


1,854 












1890 


2,094 


2,507 


1840 


212 




391 


1894 
1900 


2,242 
2,421 


3,437 


1845 


303 






1904 
1910 


2,530 
2,810 


4,769 


1850 


398 




696 


1920 


3,205 


5,621 


1854 


507 












1860 


749 




1,175 









14. Plot the population curve of New York -City. What 
has Vjeen the average rate of increase for 10-year intervals and 
for yearly intervals, approximately, since 1810? Note that 
this requires the solution of the equations 120(1 + i) w = 4769, 
and 120(1 4- i) m = 4769 ; solve by taking the logarithm of 
both sides. 

15. Discuss the increase of the population of the United 
States from 1820 to 1920 as in problem 14 the population of 
New York City is discussed. 

3. Graphical representation of algebraic functions. — To repre- 
sent a point on a given line only one number is necessary with 
a point of reference and some unit of length. To every num- 
ber corresponds one point and only one and conversely to 
every point corresponds one number and only one number. 



GRAPHICAL REPRESENTATION OF FUNCTIONS 67 



The distance cut off from 0, the origin, by any point on this 
line may be called the abscissa of the point ; a moving point 
upon this line may be designated by the variable x, which i^ 
then thought of as assuming different values, corresponding to 
the different positions of the point upon the line. 



1 1 



If another scalar line, OY, be taken intersecting OX at 
90°, the two lines may be conveniently used to represent the 
position of any point in the plane of the two lines. The two 
lines of reference are called commonly the x-axis and the y- 
axis respectively. 

The position of a point on the earth's surface is given by a 
pair of numbers representing in degrees longitude and lati- 
tude ; the -j- and — of our numbers are replaced by E. and W. 
in longitude, and by N. and S. in latitude. If we agree to give 
longitude first, then + and — could, in both terms, replace 
the letters, and position 
on the earth's surface of 
any point can be given 
by a pair of numbers. 
The system of represent- 
ing points in a plane is 
not essentially different. 

Given any point in the 
plane as P, a perpen- 



dicular is dropped to the 
horizontal line. The dis- 
tance cut off on this 
horizontal line is called 
the abscissa or ^coordi- 
nate of P; the distance 
cut off on the vertical 
line OF by a perpen- 
dicular from P to OF is called the ordinate or ?/-co6rdinate 
of the point P and it is evidently equal to the i_ PM dropped 



__. 


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Location of points in a plane 



68 UNIFIED MATHEMATICS 

to the axis OX. The two numbers together, abscissa given 
first, serve to locate the point ; thus a point P x 7 units to the 
right of OF and 5 units above OX is located on our diagram. 
To this point corresponds the pair of numbers (7, 5) (read 
" seven, five ") and to the pair of numbers (7, 5) corresponds 
point P lm The point P 4 symmetrical to P x with respect to 
OX, is (7, — 5) the negative ordinate indicating that the point 
is below the x-axis. P 2 (—7,5) and P 3 (—7, —5) are located 
upon the diagram. 

A moving (or a variable) point P in the plane is designated 
by (x, y), which is read "x, y" (not u ,x and ?/"), and the 
coordinates, abscissa and ordinate, of P are a different pair 
of numbers for each position of the point, i.e. x and y are 
variable. 

Every point represents a pair of numbers, and consequently 
a series of points will represent a series of pairs of numbers. 
In the statistical diagrams the pairs of numbers are numbers 
functionally related. In an algebraic function, y = 3 x + 5, 
we have involved a relationship corresponding to a mass of 
statistical information, and the pairs of numbers can be repre- 
sented upon a diagram just as before. Corresponding num- 
bers, a pair of numbers, are obtained by giving a value to x 
and computing the value of y. The points are seen to lie 
upon a straight line, which we shall see includes all points and 
only those points whose coordinates, abscissa and ordinate, 
when substituted for x and y, respectively, satisfy our given 
equation. This line is called the graph of the function, 3 x +5, 
or the locus of the equation, y = 3 x + 5 ; the operation of 
locating the points and connecting them is termed plotting 
the graph. 

To represent on cross-section paper any equation in two 
variables x and y, t and s, u and v, or by whatever letters 
designated, two intersecting scales as axes of reference OX 
and OY, OTand OS, or OU and OF are taken, and pairs of 
values which satisfy the functional relationship are plotted as 
above. 



GRAPHICAL REPRESENTATION OF FUNCTIONS 69 

4. Historical note. — The invention, or more properly the 
discovery, of analytical geometry was made in the early part 
of the seventeenth century. The first work directly on the 
subject was published by Rene Descartes in 1637, La Geome- 
trie, a work small in compass but great in its effect upon the 
development of mathematics and science. Almost simultane- 
ously another Frenchman, Pierre Fermat, also discovered the 
methods independently of Descartes. 

The idea of coordinates, called Cartesian after Cartesius 
(Latin form of Descartes) was not new ; in fact, as we have 
noted, this idea is found in the latitude and longitude of Hip- 
parchus (200 b.c). The idea of coordinates for drawing simi- 
lar figures was known even to the early Egyptians, and this 
idea was used for surveying purposes by Heron of Alexandria 
(c. 100 b.c). The idea of fundamental properties of any curve 
as related to its axis or axes or to tangent lines and diameters 
was also not new. The new point was to combine these ideas, 
referring several curves and straight lines to axes geomet- 
rically independent of the curves, using letters to represent 
constant and variable distances associated with the curves 
and lines involved ; the graphical representation of negative 
quantities is a vital part of the analytical geometry. These 
developments were made both by Descartes and by Fermat. 

Modern mathematics begins with this analytical geometry 
and with the calculus which was developed within a century 
after Descartes by Newton and probably independently by 
Leibniz. 

5. Industrial applications. — At the present time the graphical 
representation of statistics is playing an increasingly important 
role in many industrial enterprises. Curves derived from ob- 
servations, empirical curves, are expressed in graphical form 
for convenience of reference and, frequently, for interpolation 
between observed values. The normal distribution curve is 
employed not only by statisticians but also in the production 
departments in many factories in the classification of their 
products. 



70 



UNIFIED MATHEMATICS 



ILLUSTRATIVE EXAMPLES 

To plot a function of jt, give x values, find the corresponding 
values of y, or conversely, and plot the points. Connect by a 
smooth curve passing through all the points in succession 
moving continuously from left to right. 

1. Plot the graph of 
the function x 2 +4 x— 5, 
i.e. plot the locus of the 
equation, y=x 2 +4:X— 5. 



Give to x the values from 
to 3 and from to — 6 ; 
beyond these values in 
either direction the values 
of y evidently become very 
large. The curve is evi- 
dently symmetrical with re- 
spect to a line parallel to 
the axis and 2 units to the 
left. 

The points where this 
curve crosses the x-axis rep- 
resent solutions of the equa- 
tion x 2 -f 4x — 5 = 0. 




Points on 


the Curve 


X 


y 


-6 


+ 7 


— 5 





-4 


-5 


-3 


-8 


-2 


-9 


-1 


-8 





-5 


1 





2 


7 


3 


16 



Graph of 



x 2 + 4 x - 5 



2. Plots = 20 1+ 50. 







1 


2 


3 


4 


5 


6 


7 


8 


9 


10 


1 


1.1 


1.2 


1.3 


1.4 


1.5 


1.6 


1.7 


1.8 


1.9 


2 



Points Points 

70 

72 
74 
76 

78 
80 
82 
84 






50 


1 


1 


70 


1.1 


2 


90 


1.2 


8 


110 


1.3 


4 


130 


1.4 


5 


150 


1.5 


6 


170 


1.6 


7 


190 


1.7 


8 


210 


1.8 


9 


230 


1.9 


10 


250 


2 



Upper graph, s = 20 1 + 50 from t = to t 
Lower graph, s = 20 1 + 50 from t = 1 to t 



10, upper and left-hand 
2, lower and right-hand 



90 



scales 
scales 



GRAPHICAL REPRESENTATION OF FUNCTIONS 71 

The Lower and right-hand scales would be used it you were interested 
in the behavior of the function in the interval from t—\ to t = 2. By 
the teDfold enlargement you can read values to the third significant 
figure. 

This may represent the motion of a lx.dv which starting at a point 
50 feet from the given point of reference moves away from that point 
in a straight line at the rate of 20 feel per second. The units might be 
miles and hours, so that the speed would be given as 20 miles per hour; 
this may represent then the motion of a train. 

3. Plot y = J - 2 & - 1 S ./■ + 24 

The values of // are SO large that the figure OCCUpieS to., much • 

vertically. To obviate this difficulty on.' square on the axis of y is taken 

to represent ten units of y and one square on the x-axis 18 taken to 







M 


Y 1 II 1 ! ! h 


■ 

—si 

■l 


I 

-3 -2 




(uSjt) 

Vll- ■■;■ 

1 \ ;• 


M 


:::::: | ;::: 


V 


i 


...f.. 




-10- 

-20- 
-30- 






0*,- 


21) 






— 




± 


















::::::::::±± 




-40- 


— 


.... 






1: ; : ; 1 ; 



P01 
on 

< 1 



- 1 
-:; 

-2 

-1 



1 
2 

s 

4 



NTS 

no 

H\ ■ 

y 

-61 


88 

44 
80 

2i 

•"» 

-12 

-21 

-16 
8 

48 



Graph of y = x 3 - 2 x- - 18 x + 24 



represent one unit of x. This serves to compress or telescope the curve, 
but the essential peculiarities are preserved. In particular the points 
at which the curve crosses the SB-axis, the values of .r which make 
x 3 — 2x 2 — 18 x + 24 = 0, remain unchanged. These values, the roots 
of x 3 — 2x 2 — 18 x + 24 = 0, are seen to be — 4, 1.3, approximately, and 
4.8 approximately. 

In general an algebraic equation of this type is not likely to have a 
rational root, such as the — 4 above. 



72 UNIFIED MATHEMATICS 

4. Plot a* + y 2 - 36 = 0. 




Graph of x 1 + y 2 - 36 = 



In drawing the graph of this function of x (implicit), it is important 
to note that there are two values of y corresponding to each value of x, 
and that these two values are symmetrically distributed with respect to 
the z-axis. Similarly this curve is symmetrical with respect to the x-axis, 
since any value of y gives two corresponding values of x, numerically 
equal but opposite in algebraic sign. The points when located are con- 
nected by a smooth curve which is here a circle. 

To this diagram reference has been made in problem 7, page 17. As a 
circle of radius 6 the ordinates at x = 1, 2, 3, 4, and 5, respectively, give 
graphically the square roots of 35, 32, 27, 20, and 11. 

The more complete discussion of equations of this type is given in 
Chapter XIV. 



GRAPHICAL REPRESENTATION OF FUNCTIONS 73 

5. Plot ij = x 2 for aj = 0, .1, .2, ■•-, 1.0, 1.1, 1.2, 1.3. 

Note that precisely the same curve is obtained if units 0, 1, 2, 3, ••• 15 
are taken instead of tenths on the z-axis and tens in the place of tenths 
on the vertical axis, as indicated on the lower and right-hand scales. 













Points on t 






: : P 


OIN r 

ieC 
Rig 

AND 

Lo\ 

SCA 












the C 


;rve, 












, .. ^ j-^y 


LEiT- 




ht- IT vie 








2 i 










Upi 












,-er r- /■ 1 i.U" 








j i 1 ztzn 






-i 


LES ill 1 / 






X 

-.2 
-.1 

.1 
.2 
.3 
.4 
.5 
.6 
.7 
.8 
.9 
1 


y " 








: 4 E 








-2 

-1 

n 






:± - 


4 1 ! ■ - 1 ' ! 


01 




In i / or 


1 / yu 






11 Hi / 








ni * 




() : ! ' . j i / 




j «n 




1 


i i y °m 


.01 j_ 




i ' / 






/ 


.04 g 




< ; i / i 




4 / TA 




3 


7 /u- 


no 




o : ; i / 








• 16 ^6 






s 


If! ' <v/ fin 






5 




.25 ii 




25 i < ■■ i A / ■ zjz 


j 
.36 r 5 














7 

8 

9 

10 


36 j|| | ' £>U 












■in i i i i ! i / 






49 i I i i i i / 


64 3 


--- 


/ in 


fij. Ml 1 / w 


,o-± -r 




o-i - • Ml 1 / 








81 i 




81 !/ 


1 ±? 




/ orv 


1 „- ' / 1 ot 1 


1 




100 i :i|i /i t 
















1 !/ 


^-+Mf +2 


-p 


L ~H - " 


i^f[-r ' 1 i U^MLL±^r " ' ' ' ] ! ' ~["Tj~ p°~ 
















' i i i Lr , i M i i 






i i 1 \Jr> ! i i i i i 1 - • -t 






^ n 






IJ^'T, : ! : --- --- 1-0 






j>^< ''I Ii " " TT 








i^~^L_ 


— ^"^ - 


^ ....... 4X: 


[ -2 -1 0- 


^.1!__.2 


-+Hs!+fet 'o-fHfi- -+7- ; «- 9 -*-i--o- 1-1- - 


p> — =1 






,0 


1 2 


3 4 ; 'o : , 6 7 '«' '9" 10 z\V\ 











Graph of y — x 2 



This Une if drawn somewhat carefully can be used to read squares of 
numbers of two places to two or three places. Thus (.85) 2 = .72 ; (.96) 2 
= .92 ; (.73) 2 = .54. Square roots can also be read from this curve by 
noting the horizontal length, corresponding to any given vertical length. 
The square root of 20 is read as 4.45, of 30 as 5.50, of 40 as 6.35, of 50 as 
7.05, of 60 as 7.75, of 70 as 8.38, of 80 as 8.95, of 90 as 9.5 ; the square 
root of 630 must read as 6.3 down between 2 and 3 on the horizontal 
scale, evidently about 25. 



74 



UNIFIED MATHEMATICS 



- — - -p.-- 1 


i— 


It ± 


t 




r 




t 




t 




r 




L 


1 1 


h 


1 




19 J 1 




1 r T 




X I H f 




X t X 




X t ± 




11 - - 4- 


Ti-f.'i- 


1 1 7 




% - t 




X t 


7 x 


X - C 


x t 


i-n- - - 4- 




^ . T 




X . ] 




x - ] 




x j i 




fit - I t 




x^ t t 




- X 


i 3 x 


_ 3 - - ■ Xt 


X EX 


- ^1 


x j i 


»- i?. &••/ 


-ftO / £0(1- 


o o £f 




3T ■? " 


x -^a 




x ^? 


XJ 3t 


x J& 


r- Hi i^i 




Ml o/ 






it ~^/ X 




X ^/ X 




X X3 X 












V 




V 




' J 






r> if] r / 




/qr _5qc 


rv 2 X 


^1t Xt 


i &y X 


t ■ ^ 


/a7 


i -t -2 




-^ -j r 




7 7 


■J 


t X Z 




t -j 




i / > 




/ ' 












T r 












j 1 




1 y 




t J- * 




1- a 2 -,_ 




1 / f S 




/ t _- ? 


X 


1 _,? ***^ 


x X 


? S. _'_ 


___ ± 






:(r " ;2: :3: 3: :f 


>_ it *'- !$ w 1) I 




i r 




X i: 




______ _ ___ ±_x 



y = x z 
x 



.2 



10 



.008 

.064 

.216 

.512 

.729 

1. 

1.728 

2.744 

4.096 

5.832 

8 

27 

64 

125 

216 

343 

512 

729 

1000 



Graph of y = x 3 , using three different scales 



The portion of the cubical parabola, 

V = £ 3 , 
given by positive values of x from to 10. 

The cubes of numbers from to 10 can be read quite accurately to 
two significant figures, with an approach to the third. 

Thus (8.4) 3 is read on the 1 to 100 scale curve as 595, instead of 593 ; 
(X4) 3 is read on the 1 to 10 scale curve as 85.0 instead of 85.2 ; (1.8) 3 is 
read on the 1 to 1 scale curve as 6.00 instead of 5.83. 



GRAPHICAL REPRESENTATION OF FUNCTIONS 75 

PROBLEMS 

1. Plot y = x?, from a* = l to it* = 5, using one half inch, on 
the vertical axis to represent 10. 

2. Plot y = it 3 , from x = 1 to x = 10, using one half inch on 
the vertical axis to represent 100. See graph, above. 

3. Plot y = x 3 , from x = to x = 1 by tenths, using 10 half- 
inches for 1 unit on each axis. 

4. Read from the above curves the following cubes, to 2 
significant figures : 

3.2 3 , 4.7 3 , .82V 1.5 s , .64 3 , .58 3 , 7.1 3 , 9.2 3 . 

5. Plot the graph of s = 200 t - 16 £ 2 , from £ = 1 to t = 12. 
This represents the height at time £ of a ball thrown upward 
with a velocity of 200 feet per second. 

6. Plot the graph of s = 1000 - 200 t - 16 t 2 , from t = 1 to 
£ = 4 ; this represents the height above the earth of a ball 
thrown down from the top of the Eiffel tower with a velocity 
of 200 feet per second. 

7. Plot the graph of x 2 4- y 2 — 100 using \ inch as 1 unit on 
each axis. Find from the graph ten pairs of numbers whose 
squares summed equal 100. 

8. The area of a circle is given by the formula A = irr 2 ; 
plot the graph of the function A from r = to r = 10 inches ; 
use 3.14 for ir. 

9. The capacity in gallons of a cylindrical can of height 10 
inches having a diameter of d inches is given, within t l of 
1 % , by the formula : 

C= -J-g- • d 2 + j-Jqq- • d 2 . 
Plot the graph for d = 1 to 20 and interpret as gallons per 
inch of height. How could you find from the graph the ca- 
pacity of a can having a diameter- of 8 inches and a height of 
9 inches ? Check by cubic inches, using 231 cubic inches to 
the gallon. 



CHAPTER V 

THE LINEAR AND QUADRATIC FUNCTIONS OF ONE 

VARIABLE 



1. Theorem. — Any equation of the first degree in tivo vari- 
ables (x and y) has for its graph a straight line. 

Proof — The general equation of the first degree may be 
written Ax + By -f- C = 0. This equation can always be put 
in one of the four forms : 



x = k, if B = 0, or if B = and (7=0; 

y = k, if ^4 = 0, or if ^L = 0and (7=0; 

y = mx, if (7=0, 

or y = mx + k. 



x = k represents a straight 
line parallel to the ?/-axis, at 
a distance k units from it ; on 
such a line the abscissa of any 
point is constant. The coordi- 
nates of any point on the line 
satisfy the equation, and any 
point whose coordinates satisfy 
the equation lies upon the line. 
Thus, x = — 2 or .t + 2 = repre- 
sents a straight line, parallel to 
.and 2 units to the left of the 
2/-axis. Similarly, y = k repre- 
sents a straight line parallel to 
the #-axis and at a distance of 
k units from it. 

76 



£5=> 



I- 



-:i 



aid 



Graph of x + 2 = 
Graph of x - k = 



THE LINEAR AND QUADRATIC FUNCTIONS 77 



y = mx. Assume different points which satisfy this rela- 
tion ; the origin lies upon the locus ; (x 1} y x ) and (x 2 , y 2 ) satisfy, 

if y x — mxi, y 2 = mx 2 . 

Any two points (x 1} y,), (x£ y 2 ) 
which satisfy this equation can be 
shown to lie upon the straight line 
connecting either one with the 
origin, which evidently satisfies 
the equation. 

Consider first m to be positive ; 
the point (af 1? ?/i) may be taken in 
| y^| | the first quadrant. 

Since y 1 = mx u and y 2 = mx 2 , 



1 \~ 


i -L x 


i i 


^t 2 


t _ _ _ _r __ 


. / 


Pj. V 


1 V (ft M ? -') 




Wj! 








a/ J 




i> i "' 


i i > M ! '1/ i 












/ ' ' 






1 i / V ? 


\\\ • / 1 




















PM>' 3 1) 3 ) 
















/ 


/ , 




1 MM 1 


MM MM MM 1 



y\ = ih 



m. 



The right triangle containing jc, 
and y 1 as the sides is similar to 
the right triangle with x 2 and y 2 



Graph of y = mx 



as sides, since 



?/i 



^ • Hence 

x 2 

the corresponding angles at are equal and the points 
( x i> V\) an( i (x 2 , y 2 ) lie upon a straight line through the origin. 

Note. — If y 2 and x 2 are negative, — is in truth to be replaced by ~ ^ 2 ? 

x 2 — x 2 

since only positive quantities are involved in plane geometry. 



Conversely any point (x, y) which lies upon the given line 
drawn satisfies the given equation. For, by similar triangles, 

- = — = m, whence y = mx. 
x x l 

When m is negative, the argument is slightly changed, since 
any point (xj, y x ) which satisfies the equation must have coordi- 
nates opposite in sign ; then -&- or — S3 equals — m. 

The value m represents the rate of change of y compared with 
the rate of change of x of a point (x, y) moving on the line. 



78 



UNIFIED MATHEMATICS 



In the equation y = mx + k for any values of x, the corre- 
sponding values of y are greater by k than the corresponding 
ordinates of y = mx. Construct any three such ordinates of 
y = mx + k. Since these exten- 
sions are parallel and equal in 
length, parallelograms are formed ; 
the inclined sides of these paral- 
lelograms are parallel to the line 
y = mx and consequently to each 
other. Since any two of these 
parallel lines have a point in com- 
mon, they coincide and form one 
straight line (by plane geometry). 
The value m is called the slope of 
the line y = mx -f k, and evidentty 
varies as the angle which the line 
makes with the o>axis varies. The 
angle which a line makes with the 
sc-axis is termed the slope angle of 
the line. 

Conclusion. — Since every equa- 
tion of the first degree, 

Ax + By 4- C = 0, 
can be put into one of four forms, mentioned, every equation 
of the first degree represents a straight line. 

Conversely, every straight line is represented by an equa- 
tion of the first degree ; if the line is parallel to one of the 
axes, the form of the equation is evidently x = k or y = k ; 
if the line passes through the origin, the form is y = mx; and 
every point on the line can be shown to satisfy this equation, 
for let (a?!, y ± ) be any fixed point on the line and (x, y) any 
point whatever on the line, then 

y± = V- by similar triangles, whence 

Xi x 

l 
x 




m. 



a fixed value, and y = mx. 



THE LINEAR AND QUADRATIC FUNCTIONS 79 

Any other line will be parallel to a line through the origin, 
and its points will satisfy an equation of the form 

y = mx -f- k. 

On the line y = mx -f- k, if at any point the value of x is 
increased by one unit, the value of y, the function of x, is 
increased by m units ; on this line everywhere y increases m 
times as fast as x; the ratio of the increase of y to the 
unit increase of x, m, gives on the straight line the rate 
of change of the function y as compared with the rate of 
change of x. 



2. Intersection of graphs. — 

Plot 2 x + 3 y - 26 = 0, 

x + y — 5 = o. 

Every point on the first line is such that its coordinates 
(x, y) when substituted in 2 x + 3 y — 26 = 0, satisfy the 

equation; there are an 
>s Ev 1 1 1 1 1 FFF infinite number of such 

points, e.g. (0, -\*-), (1, 8), 
¥-), (13, 0), (¥, 1), 
1, -^ 




(2 

(-1,¥), .-. By substi- 
tuting or 1 or 2 or — 1, 
— 2, •••, for x and solving 
for y, or conversely, points 
are obtained whose coor- 
dinates satisfy the given 
equation ; similarly every 
point on the second line 
is such that its coordinates 
satisfy the equation, 

x + y — 5 = 0; 

the point of intersection 
satisfies both equations, and its coordinates can be obtained by 
solving the two equations as simultaneous. The argument is 



Graphs 



80 



UNIFIED MATHEMATICS 



entirely similar for the points of intersection of any two loci, 
representing algebraic equations ; the points of intersection 
satisfy both equations, and give a graphical method of ap- 
proximating the solutions of the equations regarded as simul- 
taneous. 

To review this demonstration, answer the - questions below, 
and read the discussion. 

What is true concerning the coordinates of every point on 
the first line ? on the second line ? What is true concerning 
the point of intersection so far as the two given equations are 
concerned? The drawing shows that (— 11, 16) satisfies both 
the equations, and substitution shows that this is precisely 
correct. In general the graphical solution is only approxi- 
mate, the degree of accuracy depending upon the accuracy of 
the drawing and the scale used. 

The point of intersection of two straight lines represents graph- 
ically the solution obtained by solving the two equations as 
simultaneous. 




\x 2 + , 



\*~Jf-n 



Intersections of y = 3 a?, and x 2 + y 2 = 25. 

The graphical presentation shows very plainly that the 
solution is, approximately, 

a; = 1.6 
and y = 4.8. 



THE LINEAR AND QUADRATIC FUNCTIONS 81 
Graphical Solution of Simultaneous Equations 









::::::::x::"::::t^""""": 




^_ 


— 




___;-_ „ , h-p-3-5 1 - 


^ — 


— - E§= 




11 N — Be- 


iiitil 




_t_ — 




iiiniiii;iiiiip 






■ ' : : ! : ; , : ' , i 
■■■■;_ ; — H^ 


1M 


1 | 1 i : i ! 




_ 20- ■ 


:::::::: ^::::::::: 






{jin — ^ ■- — j 


1*11 

EE^EEEEEEEEEEEEEEEEE 






— \— — i — r-r— i() — \%>f —^ 


t/- — 


















Q 


' : ^VJ^^' : ~i r ^ ^ 


£ it :j 5 == J 


, 1 . .1. ,■ 


1 1 Wffl i 1 1 1 1 ' 1 -4+ -4--4 ul 


±::::::::x:::::::± 



Graphs 



\y - 3x+5 



Intersections ofy = x 2 , y = 3 jc + 5. 

The graph shows that there are two solutions ; in the one, 

* = -1.2, y = + 1.4, 
and in the other, x = 4.2, ?/ = 17.8. 

These are approximate values. 

Plot carefully the graphs of the preceding problems, check- 
ing on the work presented by the graphs. 

Plot carefully these two lines and verify the statements 
made : 

2 x + 3 y — 26 = 0, j Graphically, parallel ; 
2 x + 3 y — 8 = 0. j algebraically, no solution. 

The point of intersection of two graphs represents graphically 
the solution of the two equations regarded as simultaneous. 



82 UNIFIED MATHEMATICS 

PROBLEMS 
1. Solve 

y — Sx — 5 = 0, 

3z+2a-7 = 0, 
both graphically and algebraically. 
j 2. Plot the graphs of 

y -3x -5 = 0, 
3y + 2z-7 = 0, 

Do these three lines appear to meet in one point on your 
diagram? Have these three equations a common solution? 
Substitute the solution of the first pair (obtained in problem 1) 
in the third equation. Later it will be shown that a point 
whose coordinates when substituted in a first-degree expression 
give a small numerical value is near the straight line repre- 
sented by the equation formed by putting that expression 
equal to zero. 

3. Plot 15 points whose coordinates satisfy the equation 

2y-\-3x-ll = 0. 

4. Plot the lines x = 3 and y = 4 ; what point is repre- 
sented by these equations ? Note that the Cartesian system 
(x, y) of representing points implies each point as the inter- 
section of two lines. 

5. Solve 

2y + 3x-5 = 0, 
Sy-4:X-S = 0, 

both graphically and algebraically. 

3. Intercepts. — Any given line or curve cuts off on the co- 
ordinate axes distances that are called the intercepts of the 
line or curve. The sc-intercept is obtained analytically by sub- 



THE LINEAR AND QUADRATIC FUNCTIONS 83 

stituting y = and solving, i.e. by solving as simultaneous the 
equations of the ic-axis and the given line ; the ^/-intercept is 
obtained by substituting x = 0. 

The a'-intercept of 2 x 4- 3 y — 26 = is 13, obtaining by 
substituting y = in 2 x + 3 y — 26 = 0; the ?/-intercept is 
+ ^ 6 - ; of x 2 + y 2 = 25, the aj-mtercepts are ± 5, the y-mtei- 
cepts are also ± o. 

Note that the problem of finding the intercepts of a given 
graph is a special case of the problem to find the intersections 
of two given curves ; the ^-intercept designates the inter- 
section of the given curve with the z-axis, y = 0, and similarly 
the ?/-intercept refers to the intersection with x = 0. 

Rule. — To find the ^intercept, put y = 0, and solve ; 
similarly for the ^/-intercept. 

4. Pencil of lines. — The straight lines which pass through 
a common point constitute what is termed a pencil of lines. 
If the common point is determined as the intersection of two 
given lines, we may write the equation of the pencil of lines 
in terms of the two expressions which put equal to zero repre- 
sent the given lines. 

The pencil of lines through the intersection of 

^-3^-5 = (y 

3?/ + 2^ + 7 = (Z 2 ) 

is given by the linear equation, 7c being assumed constant, 
(3) y _ s x _ 5 + fe (3^ + 2x + 7) = 0. (y 

Evidently any point on the first line, l A , makes y — 3x — 5 = 0, 
and any point on the second line, l 2 , makes 3y-\-2x-\-7 = 0; 
the point of intersection substituted in our equation (3) gives 
+ k • or 0, hence the point of intersection of l x and l 2 
satisfies equation (3) for all values of k. 

By giving k successive values l z can be made to pass through 
any point of the plane. Thus to pass through (1, 5) sub- 
stitute (1, 5) in Z 3 and solve for k, giving 

5 _ 3 _ 5 + fc(15 + 2 + 7) = 0, 



84 UNIFIED MATHEMATICS 

or 24 k = 3, k = i The line ?/ - 3 a - 5 + i(3 y + 2 cc + 7)= 0, 
or 82/-24x-40+3?/+2a;+7 = reduces to lly-22a;-33=0, 
oi y — 2x — 3 = when simplified. 

In solving as simultaneous the two equations y — 3x— 5=0, 
and 3y + 2x + 7 = 0, the particular lines parallel to the axes 
of reference and passing through the point of intersection of 
Zj and l 2 are sought. Thus, after multiplying the upper 
expression by — 3 and adding, you get the line l 3 with k = — J. 
2/ _3a;_5_i(3 2/ 4.2x + 7)=0 

gives — 11 x — 22 = 0, or x = — 2. 

To eliminate x we multiply the upper expression by 2 and 
the lower by 3 and add ; this gives, 11 y + 11 = 0, or y = — 1. 
The same line given by 11 y -f- 11 = is obtained from line 
Z 3 with k = f ; i.e. : 
2/ _3a_5_j_!(3 2/ + 2a; + 7)=Q 

gives 11 y -f 11 = 0, or y + 1 = 0. 

The point of intersection of the two lines, ( — 2, — 1), is 
given as the intersection of x = — 2 and y = — 1. 

PROBLEMS 

1. Write the equation of the family of lines through the 
intersection of the two lines : 

^-3^-5 = 0, 
3y + 2x-7 = 0. 

Determine k so that this line shall pass throught the point 
(0, 0) ; through (1, 5). 

2. Find the sc-intercept and the ?/-intercept of the line, 

3y + 2x- 7 = 0. 
Draw the graph of this line. 

3. Write the equation of the family of lines passing through 
the point of intersection of the lines, 

2/-4 = 0, 
and x — 3 = 0. 



THE LINEAR AND QUADRATIC FUNCTIONS 85 

Draw the graph of y — 4 + k(x — 3) = 0, for k = 0, for fc=l, 
for k = - 1, for k = 3. 

4. Plot the three lines below, and obtain their three points 
of intersection graphically and algebraically : 

y - x — 5 = 0, 
3y + 2a;-10 = 0, 

2/ — 3 a; + 15 = 0. 

Does the drawing on coordinate paper give an indication of 
the area of the triangle formed ? 

5. Write the equation, ?>y -\-2x- 10 = 0, in the form, 
y = mx 4- k ; what is the value of m ? 

6. The weight of a cylindrical vessel of water when filled 
to a height of 10 inches is 6.8 pounds, when filled to a height 
of 6 inches it is 4.4 pounds ; plot the two points (6, 4.4) and 
(10, 6.8). The straight line joining these two points gives the 
weight of the vessel when filled to any height from to 10. 
The equation may be written iv = k ■ 7i -f c, where iv and h are 
the variable weight and height, k and c are constants. This 
equation is the simple statement of the fact that the weight of 
the water and the container for any height h is the weight of 
the vessel (c) plus h, the height, times the weight of the water 
which fills the container to a height of one inch. Note the 
significance of the intercepts. 

7. The equation, w = -—^—y, may be used to express the 

relation between the volume in cubic inches and the weight in 
pounds of a given mass of water. Plot this carefully and find 
approximately the weights of 100 cubic inches, 500 cubic inches, 
and 700 cubic inches of water. Find the volume of 15 pounds 
of water ; the volume of 25 pounds ; of 30 pounds. 

8. The volume of mercury at any temperature between 
and 40° C. is given by the equation V=k(l -f- at), wherein 
a = .00018 ; f or k = 1000 cu. cm. this becomes V= 1000 + .18 I, 



86 UNIFIED MATHEMATICS 

Plot this equation taking the horizontal axis as at 1000. This 
is equivalent to plotting the increase in volume, I = .18 1. 
Plot for to 40° C. and find the increase in volume when 1000 
cu. cm. of mercury at 0° C. are heated to 27° C. 

9. Find the equation of the straight line through (—3, 5) 
and through the intersection of 3 a? — y — 7 = and 

5 x + 12 y - 17 = 0. 

10. Plot degrees Fahrenheit as abscissas and degrees Centi- 
grade as ordinates, connecting (32° F., 0° C.) to (212° F., 100° C), 
by a straight line. Find the equation of this straight line. 
Find the Centigrade reading corresponding to 0° Fahrenheit, 
to 100° F. Discuss the meaning of the slope of the line. 

11. Find the intercepts of the line 9 y — 5 x = — 160. Com- 
pare with your result in the preceding problem. 

12. Plot the graph of s = 16 t 2 , for values of t from t = to 
t = 5, using one inch for 1 second on the horizontal axis, and 
1 inch for 100 feet on the vertical axis. Find value of s when 
t = 4.3 from the graph. Check by computation. 

13. Plot carefully x 2 +y 2 = 64, and y = 3 x - 5. From the 
graph get the approximate solution. 

14. Show graphically how to change a system of marks from 
a scale of 100 to a scale of 75 ; from 75 to 100. 

15. Sound travels at the rate of 1089 feet per second in air 
at 32° F. (or 0° C.) ; at the rate of 1130 feet per second in air 

13 t 

at 70° F. The formula, v = 1054 -f- — — gives very closely the 

velocity in feet per second at temperature f Fahrenheit. 
Plot the graph of the function, plotting the excess above 1000 
feet as ordinates and temperature Fahrenheit up to 80° F. as 
abscissas. At what temperature is the velocity 1100 feet per 
second? How would you adapt these figures to the Centigrade 
scale for temperature beginning 0° C? v = 1089 + 2 t is the 
resulting equation. 



THE LINEAR AND QUADRATIC FUNCTIONS 87 

16. The velocity after t seconds of a bullet shot straight 
upwards at 800 feet per second is given by the equation 
v = 800 - 32 t. Plot the graph, taking 100 feet as i inch 
on the vertical axis, and o seconds as i inch on the horizontal 
axis ; negative values of v mean that the bullet is descending. 

17. Plot v = 600 + 32 1, and interpret as downward velocity 
of an object thrown downwards from a height. 

18. Time yourself on plotting the following 10 lines ; five 
may be plotted with respect to one set of axes : . 

a. 3x + 4y-12 = 0. V-l^l 

b. 3y = 2x-5. y ' 3 5 

d. 7<b + 32/-18 = 0. l x =iy~i- 

e . 2y + x 4-10 = 0. J- Sx-7y = 0. 
/. 5a?4- 12y — 10 = 0. 

19. Time yourself (a) on finding the slope of each of the 
ten lines in problem 18 ; (6) on finding the ^-intercept of 
each line ; (c) on finding the ordinate of the point whose 
abscissa is 2 ; (d) on finding to one decimal place the ordinate 
of the point on each line whose abscissa is 2.4 ; (e) on putting 
these lines in slope form. 

20. Plot, using values correct to 1 decimal place, the follow- 
ing lines : 

a. 3.1x + ±.5y-12 = 0. 

b. 3.2 y = 2.6 x- 5.7. 

c. .93-4.8^-8.3 = 0. 

5. The quadratic equation in one variable. — Any equation of 
the form ax 4- b = is called a linear, or first-degree equa- 
tion, in the variable x ; the solution is given by x = ; the 

a 

graph of the function, y = ax -\- b, is a straight line of slope a, 
with the y intercept equal to b, and with the x intercept repre- 
senting the solution of the equation, ax + b = 0. 

Any equation of the form ax 2 + bx -f c = is called a quad- 



88 UNIFIED MATHEMATICS 

ratic equation in x; a, b, and c are to be regarded as con- 
stants. The graphical solution of one equation of this type 
has been presented (page 70) and the algebraic solution is 
given in elementary algebra, but will be given here a rapid 
review. 

Algebraical solution of 2 x 2 -f- 8 x + 7 = and of the general 
equation, ax 2 -\- bx -+- c = ; 

2 x 2 + 8 x + 7 = 0, ax 2 + bx + c = 0, 

x 2 + 4x = -i x 2 +^ = 



a a 



x2 + 4 * + 4 = i x 2 + h -x+(±^ = ]*-- c ~ 

2 ' o V2a/ 4 a 2 a 



/ ^ \ 2 52 _ 4 



6 \ 2 6 2 - 4 ac 
2 ' 



7) _i_ V7>2 4. r»/» 

a; + 2 = ±.71 (or .707 to 3 places), «; + -?- = - . 

v J 2 a 2 a 

-. on o r~t —b± Vb 2 — 4 ac 
x = — 1.29 or —2.71, a; = =-^ . 

2 a 



The necessary third term to complete the square is obtained 
by comparison with (x ± k) 2 = x 2 ± 2 kx 4- k 2 . 

Graphically the equation 2/ = 2sc 2 -f8cc-|-7 represents a curve 
which intersects the sc-axis, y = 0, in the two points whose 
abscissas satisfy the equation, 2 x 2 -\- 8<c+7 = 0. 2/ = 2x 2 +8a; + 8 
represents a curve which is tangent to the a>axis, correspond- 
ing to the fact that the roots of the equation, 2 x 2 -\- 8 # + 8 = 0, 
are equal to each other. The equation y = 2x? + %x + ll repre- 
sents a curve which does not cut the sc-axis, corresponding to 
the fact that the quadratic 2x 2 + 8x + ll = has for solu- 
tions, x = = — = — ^— , values corresponding to no points on 
A 

the cc-axis, i. e., to imaginary values of x. Plot the graphs in- 
dicated. 



THE LINEAR AND QUADRATIC FUNCTIONS 89 

The quadratic equation is solved algebraically by reducing 
the problem to the solution of two first-degree equations : 

■ & 



+ V^= 


- 4ac 


o 


i 
a 


-V6 2 - 


- 4 ac 



and x + — = 

2 a 2 a 

The quantity b 2 — 4 ac which appears under the radical sign is 
called the discriminant of the quadratic. The nature of the 
roots of the quadratic equation is determined by this dis- 
criminant, when a, b, c represent real quantities, i.e., a, b, and 
c having values which can be represented by points upon a 
scalar line. When 

b 2 — 4 ac > 0, i.e. positive, the two roots are real and un- 
equal, when 
b 2 — 4 ac = 0, the roots are real and equal, and when 
6 2 — 4 ac < 0, i.e. negative, the roots are imaginary. 

Further, the condition that the roots of the quadratic should be 
equal given by b 2 — 4 ac — 0, may be obtained by inspection, 
or by actually setting the two roots equal to each other and 

simplifying ; ax 2 4- bx + c may then be written a[ x + — ] • 

V 2 a) 

Graphically these conditions correspond to the fact that the 
curve y = ax 2 + bx + c cuts the a^axis in two points, is tan- 
gent to the a>axis, or does not intersect it at all, according as 
b 2 — 4 ac is greater than, equal to, or less than 0. 

Frequently the two roots of the quadratic ax 2 -f- bx -f c = 
are designated by x x and x 2 . 

Thus 



x l = 


_ b + Vb 2 - 


- 4ac 


2 a 


j 


Xo — — 


-b- V6 2 - 


- 4ac 



and 

za 

The sum and the product of the roots, x 1 + x 2 and sc^, are 

b c 

given, respectively, by and +-. The expressions x 1 + x 2 

and x l x 2 are representative symmetric functions of the roots of 



90 UNIFIED MATHEMATICS 

a quadratic function of one variable, being expressions which re- 
main unchanged when Xj and x 2 are interchanged. 

6. Historical note. — The solution of linear equations was 
known four thousand years ago to ancient Egyptians. The 

equation x + - = 19, was proposed and solved in the work of an 

Egyptian writer named Ahmes ; the problem reads, with " ahau " 
representing " heap " or " unknown," " ahau and its seventh, 
it makes 19." In other ancient Egyptian documents problems 
leading to pure quadratics are found. The Greeks were able 
to give as early as 450 e.g. a geometrical solution of any quad- 
ratic having positive roots ; the numerical application appears 
in Greece somewhat later. In India numerical quadratics 
were solved in the fifth and sixth centuries a.d. The first 
systematic treatise combining clearly analytical statement with 
geometrical illustration is given by an Arab, Mohammed ibn 
Musa al-Khowarizmi, about 825 a.d. His work continued in 
use for centuries. The complete quadratic with general, 
literal coefficients, did not come, of course, until after the in- 
troduction of literal coefficients by Viete late in the sixteenth 
century. 

7. Graphical solution of the general quadratic equation. — The 

general quadratic equation ax 2 + bx -f- c = can be solved 
graphically by means of one fixed curved line, y =x 2 , and a 
variable straight line. The intersection of 

y = x 2 
and ay + bx + c = 

gives the solution of the equation ax 2 + bx + c = 0, for the 
solution is obtained algebraically by substituting for y its 
value x 2 in ay + bx -f c = 0, giving ax 2 -f- bx + c = 0. 

The graphical solution of the quadratics, 2 x 2 — 6 x— 5 = 0, 
2x 2 -6x = 0, 2x 2 - 6a; + f = 0, and 2a; 2 - 6x+ 10 = 0, 
is presented upon the diagram ; the student is urged to 
solve these equations algebraically and to trace the correspond- 



THE LINEAR AND QUADRATIC FUNCTIONS 91 




Graphical solution of quadratics 

2x 2 -6x-5 = 0; 2x 2 -6x = 0; 2x 2 -6x + § = 0; 2x 2 -6x+10 = 



y=x t i 

2y-Qx-5=0. 
Two real solutions. 



2y-6x + ± = 0. 
Two coincident solutions. 



2y-6x = 0. 
Two real solutions. 



y=x 2 , 
4?/-6.t+ 10 = 0. 
Two imaginary solutions 



92 UNIFIED MATHEMATICS 

ence between the algebraic and graphical solutions. Two 
sets of real and different roots are indicated by two of these 
straight lines on our diagram; one set of equal roots is in- 
dicated ; one pair of imaginary roots is indicated by the line 
which does not meet the curve. 

8. The quadratic function. — The expression, ax 1 -4- bx -f- c, 

assumes different values when different values are assigned 
to the variable x. The variation in value of the function, 
ax 1 -+- bx + c, for given values of a, b, and c, is most easily 
given by drawing the graph of 

y = ax 1 + bx + c. 

The maximum or minimum value of the function, ax 2 -4- bx -f c, 
for any real value of x, may be found by solving, ax 2 + bx+c=y, 
and determining whether there is any greatest or least value 
which you have for real values of x. 

Thus, y = 2x 2 -6x- 5, 

2ac 2 — 6x — (5 + y)=0, 



6±V36 + 8(5 + y) _ 6±V7 6 + 8y 
4 4 - 

If y is less than — - 7 /, the values of x become imaginary ; consequently 
— - 7 / is the minimum value which y can have. 

If ax 2 + bx + c = has real roots, and a is positive, the func- 
tion ax 2 + bx -f- c is negative for values of x between the two 
roots and positive for all other values ; if the equation has 
imaginary roots, a being positive, ax 2 + bx + c is positive for 
all real values of x. 

a(x — Xy) (x — sc 2 ) is positive for values of x greater than both X\ and 
»2» negative for values of x between Xi and X2, and zero for x = X\ or 
x = x<i ; X\ and x 2 are supposed to be real, and a positive. 

is positive, a being positive, when 4ac > b 2 . 



THE LINEAR AND QUADRATIC FUNCTIONS 93 
9. Summary. ax 2 4- bx + c = 0. 



_ & 4- V & - - 4 <zc _-&-V&--4 



X l — « ) X 2 



ac 



9 ' 2a 

c 



x l + x 2 — , XiX 2 — 

a a 

Given a, b, and c, real numbers, the condition : 

b 1 — 4 ac > 0, gives real and unequal roots. 

If a, 6, and c are rational, and b 2 — 4 ac a perfect square, then the 
roots are also rational. 

b 2 — 4 ac = 0, roots are real and equal. 
b 2 — 4 ac < 0, roots are imaginary. 

PROBLEMS 

1. Solve by completing the square : 

a. x 2 -4a;-5 = 0; 6. 2z 2 4-3o;- 5 = 0; c. 3z*-4z-9 = 0. 

2. Solve by formula : 

a. 3a 2 -4#-9 = 0; 6. 7z--3z-10=0; c. 3y"<+2y-5 = 0. 

3. Time yourself in solving the following 10 quadratics, 
writing the roots in simplest form but not approximating the 
square root. 

a. 2a 2 4-3x-5 = 0. /. 9^ = 20:r + 10. 

b. 3a?-2a; + 7 = 0. g. 7P = 2£t-5. 

c. 5^ + 12^4-3 = 0. h. *2 + 4* = l. 

d. yt-Sy — 7 = 0. t. 9^ = 16-24y. 

e. 2v 2 -10u-35 = 0. j. 8m4-5 = 3m 2 . 

4. Time yourself in finding to one decimal place the roots 
in the above 10 equations. 

5. Find the nature of the roots, without completely solving, 
in the following equations : 

a. x 2 4- 3 x - 5 = 0. c. aP + 3a — 40 = 0. 

6. x 2 + 3x - 8 = 0. c?. x 2 - 3x 4- 40 = 0. 

e. x 2 - 3 x 4- 1 = 0. 

/. 4x 2 - 12 a + 9 = 0. • 



94 UNIFIED MATHEMATICS 

6. The velocity of a freely falling body is given by the 
formula, v = 32 t, when falling from rest ; or v = 32 1 + k, 
where k represents the velocity at the instant when t = 0, or k 
is the velocity at the instant when you begin to measure the 
time. Plot for values of t from to 10. 

7. A bullet shot straight up into the air at a velocity of 
1000 feet per second has its height above the earth given by 
the equation h = 1000 1 — 16 t 2 -. Plot this equation for values 
of t increasing by intervals of 5 seconds from t = to t = 70. 
If the bullet is shot at an angle in such a way that the vertical 
velocity when leaving the gun is 1000 feet per second, the 
given equation continues to hold for the height of the bullet 
above the earth. The resistance of the air (considerable at the 
velocity mentioned) is neglected in these equations. 

8. Plot the graph of y = 3 x — 7 ; give to x the integral 
values from — 2 to +5 and find the values corresponding of y. 

9. A freely falling body falls from rest in t seconds a dis- 
tance s, given by s = 16 1 2 ; plot points given by corresponding 
values, using horizontal axis as £-axis, and vertical axis for 
distance. Take values of t from to 10, and as .9 will vary 
from to 1600 take 1 cm. to represent 100 on the s-axis. 

10. For what values of x are the following expressions : 
a. 0-3)0-5); b. (x + 1) (x - 4) ; c. 2x 2 + 3x-5; 
d. 4^-12x + 9; e. 4.x- 2 - 8x + 9; /. 3« 2 + 2x- 7. 

11. Given h=S00t-16t 2 , find t when ft = 100, 1000, 
10,000, 12,000 respectively. This equation represents the 
height to which a bullet would rise when shot vertically up- 
wards at a velocit}' of 800 feet per second, neglecting air- 
resistance. Interpret your results. This bullet has a velocity 
at time t, v = 800 — 32 1 ; find the velocity at the various 
heights mentioned. 

12. Solve 16 1 2 — 800 1 + h = f or t, regarding h as a con- 
stant. See preceding problem and find maximum value h can 
have. 



THE LINEAR AND QUADRATIC FUNCTIONS 95 

13. In solving 16 t 2 — 800 1 + h = 0, two roots are obtained ; 
find the sum of these roots and the product. Interpret the 
sum, i.e. give the physical meaning. 

14. Discuss the changes in value of 2>x 2 + 2x — 5 as x 
changes from — 10 to — f , to 0, to 1, to 2, to 10, to " positive 
infinity." 

15. Determine the nature of the roots : 

a. 4^-16^-160 = 0. c. 3v 2 + 16 v + 20 = 0. 

b. 7^ + 16* -160 = 0. d. 3v 2 + 16 v + 25 = 0. 
•16. Plot the graphs of the functions in 15. 

17. Solve the equations of 15 graphically, using the inter- 
section with y = t l OT y — v 2 (one half -inch may be taken for 
10 units on the vertical axis). 

18. Find the sum and the product of the roots in the 
problems of 14 and 15. 

19. Solve I a?2+ y2 = 36 > by substitution. 

[y = 3x + 5, 

20. Solve \ ~ ' by substitution. Draw graphs. 

[ S ^ ot — t), 

21. Solve .1 1 2 - 50 1 - 30 = 0, to 2 places of decimals. 

22. Solve t 2 — 50 1 — .0001 = 0, to 2 places of decimals. 

23. Find the maximum or minimum values of the following 
quadratic functions : 

a. x 2 — 4x + 4. c. x 2 — 4# + 6. 

b. 2z 2 + 3:e + 5. d. a 2 + 4. 

24. For what values of x is x 2 — 6 # — 16 positive ? 

10. Equations reducible to quadratics. — The solution of 
ax 2 -f bx + c = 
is a value of the variable x, which when it is substituted in 
ax 2 + bx + c, makes the expression ; similarly this solution 
gives a value of the variable v, or £, or p, or f 2 , or ** or 3 1 2 — 1, 
or 7 £2 _j_ 2 £ — 3, which makes the expression of the same form 



96 UNIFIED MATHEMATICS 

iii that variable zero ; viz., a value which makes av 2 + bv + c 
equal zero when the value is put for v, or a(t 2 ) 2 + bt 2 -f- c, equal 
zero when the value is put for f, or a(3 t 2 — l) 2 -f 6(3 t 2 — 1) + c 
equal zero when the value is put for 3 1 2 — 1. Equations 
which can be put in the form ax 2 + foe + c = are called 
equations in quadratic form, the term being applied, in gen- 
eral, to expressions which are not quadratics directly in the 
principal variable. Thus, in any expression involving x, x 2 , x 3 , 
x A , x 5 , or x 6 , the value of the expression depends primarily upon 
the principal variable, x ; an expression like 3 sc 4 — 2 x 2 — 7, 
involving the variable x 2 , its square, and constants as coeffi- 
cients, is said to be in quadratic form, and it is a quadratic in 
the variable x 2 , but a quartic in x. 

9. Illustrative exercises. 

1. Solve St 4 -5£ 2 -7 = 0. 

As a quadratic in t 2 , the formula for the solution of a quadratic gives 



t2 = 5±V25 + 84 = 5±V109, 
6 6 



>±V109. 
y 6 
There are four values represented here, of which two are imaginary. 

2. Solve x z = 1, or x 3 — 1 = 0, and x? — 8 = ; these illus- 
trate a type of equation reducible to a quadratic by factoring. 

X3 - 1 = {X - 1) (X2 + X + 1) =0. 

a - 1 = 0, x = 1 

^ + x+l = 0, g = -l±Vl-4 = -l ± y=8. 

2 2 

These values ^ — — — ^— , ^ — ^ — ^— and 1 are called the cube 
2 2 

roots of unity ; note that V— 3 is denned as a quantity whose square is 

— 3 ; the systematic discussion of such numbers is deferred until a later 

chapter. Squaring either of the two imaginary cube roots of unity gives 

the other ; these roots may then be designated as 1, w, w 2 . The cube 

roots of 8 are 2, 2 w, and 2 w 2 ; of 7 are 7 3 , 7 ¥ w, and 7 3 w 2 , wherein 7 5 

denotes the real cube root of 7. 



THE LINEAR AND QUADRATIC FUNCTIONS 97 

PROBLEMS 

1. Solve for t\ and then for t. t* - 7 t 3 - 8 = 0. 

2. Solve and check by substitution : 

ar 4 + 3 ar 2 - 5 = 0. 

3. 2 at— 7 a;* -5 = 0. 



*• H) ,+ H)- i - a 



Note that this expression when cleared of fractions gives 

X 4 + £3 + X 1 _|_ x + 1 - 0, 

a factor of x 5 — 1 = ; the imaginary roots of x which are obtained by 
solving are the other four fifth-roots of unity. 

5. (3 a* - 5) 2 + 2(3 x 2 - 5) - 7 = 0. 

6. v + v* = 10. 

* 2 -l * 

8. Find the value of a;^, and of x in 

a; -f 3 vfr - 7 = 0. 

9. Find the value of x*, in 

x _ 3 x \ :- 7 = o, 

and compare with the preceding. The real test of a value 
found as a root is obtained by substituting the value in the 
given expressions. Squaring may introduce a new root ; thus 
squaring x = 2, gives x 2 — 4, or is equivalent to multiplying 
x — 2 = 0, member by member by x + 2. 



10. 3 a; + Va* + 5 = 7. 11. 3 a; — Va? + 5 = 7. 

10. Limiting values of a, b, c. — As c approaches more and 
more nearly to zero as compared with a and b, it is evident 
that some value of x also near to zero will satisfy the equation 
ax 2 +k-)-c=0; this value will be of the same sign as c if b is 
negative, and opposite in sign to c if b is positive. This may 



98 UNIFIED MATHEMATICS 



be obtained by taking the approximate value of V 6 2 — 4 ac as 

b (see page 24), giving and -f- - as the approxi- 

2 6 b a b 

mate values of the two roots when c (or a) is small in compari- 
son with b. 

Thus, 3 x 2 - 2 x - .000001 = 0, 



„. _ 2 ± V4 + .000012 

2 ± 2.000003 4.000003 .000003 
- = or 



6 
= | or - .0000005. 



Similarly in 1000 x 2 — 3000 x — 1 = 0, one solution will be small, 
approximately 3 -^. When c = 0, the roots of ax 2 + bx+ c = 
are the roots of ax 2 + bx = 0, giving x(ax + 6)= ; whence 

#=0 and a; = — . When both b and c approach zero, both 

a 
roots of the quadratic ax 2 -\-bx -\-c = approach zero. 

When a approaches zero in comparison with b and c, one 
root of the quadratic becomes very large and the other ap- 
proaches . Thus in the quadratic 

x 2 - 1000^-3000 = 0, 



x = 1QQ0 ± VlOOOOOO + 12000 _ 1000 ± 1006 = 1003 or _ 3 
X 2 2 

/1000 ± 1005.982 



V 2 

or - 2.991. 



are the more exact values, giving 1002.991 



As a approaches nearer and nearer to zero one root becomes 
larger and larger without limit. Thus if above we had 

.001 x 2 - 1000 x -3000 = 



_ 1 000 ± VlOOOOOO + 12 = 1000 ± 1000.006 
X ~ .002 .002 

= 1000001.5 or - 3 (more exactly 2.999991 as before). 



THE LINEAR AND QUADRATIC FUNCTIONS 99 

Both roots become large if both a and b become small as com- 
pared with c. 

PROBLEMS 

Find first approximate values, and verify by solving the 
quadratic : 

1. 3 .t 2 - 7 a -.0001=0. 

2. 5x 2 — 7x—.l = 0. 

3. 5 x 2 - .007 x - .001 = 0. 

4. 4000 = 3000 1 — 16 t 2 ; one root is the number of seconds 
for a bullet to rise 4000 feet, initial velocity 3000 feet per 
second, air resistance neglected ; what does the other root 
represent ? 

5. .01s 2 -300a;— 500 = 0. 

6. .003 t 2 + 2 1 — 42 = ; this gives a more exact equation for 
the temperature at which the velocity of sound in air becomes 
40 feet greater than it is at 0° C. 

7. 1000000 x 2 - 3000000 a T- 5 = 0. 

REVIEW PROBLEMS 

1. Plot the graph of y = 3 x — 5. 

2. Plot the graphs of the following functions : 

a. y — x 2 — Ax + 5. 

b. y = x 2 — 4 x 4- 4. 

c. y = x 2 — 4 x. 

d. y = x 2 — 4 x — 2. 

3. For what values of x is y equal to in the four functions 
of the preceding question ? The graphical solution is desired. 

4. Plot 15 points from x = — .5 to a; = + 8 and join by a 
smooth curve representing 

?/ = 2z 3 +6a;2-10a;-8; 
for what values of x is y equal to zero ? 



100 



UNIFIED MATHEMATICS 



5. s = 20Z.+ 50 — 16 1 2 . This equation represents the mo- 
tion of a body thrown from a height of 50 feet straight up into 
the air with a velocity of 20 feet per second. Plot the graph 
and locate the position of the body at the end of 1 second ; at 
the end of 5 seconds. 

6. Plot the graph of s = 800 1 — 16 1 2 , for values of t from 
to 50 ; note that it is desirable to get the values of s first for 
intermediate values and to choose the ^/-scale accordingly. 
This equation represents approximately the height after t sec- 
onds of a bullet shot straight into the air with a velocity of 

800 feet per second. 



x-axis and ?/-axis off the paper 



' ' 1 ' 


± :^ 










:r 


± 


x 














A\3\j- 










ioU J 


_L 


± 


± _r 
























"latr 


III 


11 


inn 




- - 1 2 1 2^5- - 1^3- -il-p-h- nl 4 4" ~1 J 4 L <7 -PT 




t i r fi "~m i 


_ i .. i m .in 



Shifted lines of reference 



7. Plot the graph of 

V= ^ between d = 12 and 
4 

d = 20, taking the scales so 
as to enable you to read vol- 
umes as correctly as possible 
within these limits. Plot 
only values above 100 on the 
2/-scale, and to the right of 
12 on the ^-scale. This gives 
the volume in cubic units per 
unit of height of cylindrical 
containers which have radii 
varying from 12 to 20 units. 
Apply this to cans and to 
silos. 



v*l 



8. Plot the graph of t 2 = - — : this gives the time of beat of 

a pendulum I centimeters long where gravity is 980 cm. per sec. 
per sec. 

9. Plot the graph of y — x% , for values of x from to 8. 
10. Plot the graphs of the following linear functions : 

a. y = Sx-5. c. v = 10 + 8t. e. s = 100-40*. 

&. y = 3». d. s = Q-3t, f. y^~2x + 10, 



CHAPTER VI 
STRAIGHT LINE AND TWO-POINT FORMULAS 

1. Slope-intercept formula : y = mx + k. 

The equation y == mx + k, into which, form the equation of 
any straight line can be put, is called the slope-intercept form 
of the equation of a line ; m represents the slope of the line and 
k is the intercept on the ?/-axis. The equation of a line parallel 
to the ?/-axis, x = k, cannot be placed precisely in this form, as 
the ^/-intercept is infinite. 

2. Point-slope formula : y — y x = 772(jr — x^). 

As it is frequently desired to find the equation of a line of 
given slope and passing through a given point, a separate 
equation in terms of the slope and coordinates of the given 
point is desirable. Let the equation of the line be conceived 
as in the form, y = mx + k ; since (x ly y^) is on the line, 
y l = mx 1 + k ; subtracting gives y — y 1 = m(x — a^), the equa- 
tion of the straight line in terms of m, the given slope, and 
(#1? Vi) the coordinates of the given point. 

3. Two-point formula: V~zM ^WkZlMk. 

X — X ^ x 2 — x^ 

The equation of the straight line through (x l} yi)(x 2 , y 2 ) is 
also easily derived from the slope-intercept form. 

As before y x = mx Y + k, 

y 2 = mx 2 -f- k, 
whence y 2 — ■ y± = m(x 2 — x x ), 

and m = ^ 2 ~ - yi , giving m, the slope 

a?2 — a?j_ 

of the line, in terms of <c 1} 2/u # 2 , and 2/ 2 - 

101 



102 UNIFIED MATHEMATICS 



Hence, y — y l = ^ — ^ (x — x x ) is the equation of the line 
x 2 — x x 
in a form involving only the given constants. 

The expression, m = ^ 2 ~~ ^ l , represents the slope of a line 

X 2 — x± 

joining (#!, y{) to (x 2 , 2/ 2 )- Similarly ^ ~ ^ * represents the slope 

x — x 1 

of the line joining any point (x, y) to (x 1} yi). The preceding 

equation of the line in the form ^ ~ ^ l = ^ 2 ~ •^ 1 is an equality 
ot two slopes. L A L 

The formula m = ^ — x should be memorized. 

This formula gives the rate of increase of y in the interval 
from (a?!, y x ) to (a? 2 , 2/2) as compared with the increase of x in 
the same interval ; it compares the change in y in the interval 
with the change in x in the same interval. 

If ^ = x 2 , the line joining (x ly y x ) to (x 2 , y 2 ) should be given 
directly as x = x lf parallel to the x-axis, and similarly if y± = y 2 . 



PROBLEMS 

1. Find the equation of the line of slope 3 and y-intercept 5 ; 
with m=3, Jc=— 5; m=— 3, Zc=8; 7n = 0, &=4; m=5, k=0. 

2. Put the following equations into slope-intercept form : 
a. 3 y - 2 x + 5 = 0. d. y - 3 x r- 7 = 0. 

6. 3ar+2y-7 = Q, e. y + 5 = 0. 

c. <e + 2 2/ = 0. /. a + 3 = 0. 

3. Write the equation of the straight line through (—2, 5) 
and (1, 4) ; through (3, — 5) and (2, 1). Find intercepts on 
both axes and the slope in each case. 

4. Write the equation of the straight line through (1, 5) 
having the slope 3. Find the x and y intercepts. 

5. Find the equation of the straight line through (a, 0) and 
(0, b), i.e. the line having intercepts a and b, respectively, 



STRAIGHT LINE AND TWO-POINT FORMULAS 103 

and put this equation into the form - + ^ = 1. This is called 

a b 

the intercept form of the equation of a straight line. 

6. Given 9 C = 5 F — 160, the formula connecting centi- 
grade and Fahrenheit readings of temperature, find the slope 
and the x and y intercepts. Find the slope of the line join- 
ing (32, 0) to (212, 100). What is the rate of change of G in 
the interval as compared with the change in F? What physi- 
cal meaning have the intercepts ? 

7. Given that 1000 cu. cm. of mercury at 0° C. increases to 
1007.2 cu. cm. at 40° C, find the rate of change of volume per 
degree of temperature, and finally per cu. cm. Note that it is 
not necessarily true that this rate found for an interval of 
40° C. should be the uniform rate everywhere in the interval. 
Write the equation representing the volume in terms of 
temperature, assuming that the relation is linear, i.e. that the 
increase in volume is proportional to the temperature. 
Mercury expands differently at different temperatures, but the 
variation is slight in the interval from 0° to 40°, not varying 
by more than | of 1 % from .00018 cu. cm. per degree for 1 cu. cm. 

8. Join (0, 0) to (100, 39.37) and interpret for converting 
centimeters to inches and inches to centimeters ; what is the 
meaning of the slope ? Find the value in inches of 18 cm., 
39 cm., 47 cm. Note that 100 cm. = 39.37 inches. 

9. 59.8 pints of water weigh approximately 62.4 lb. Draw 
the graph connecting (0, 0) to (59.8, 62.4) which will give the 
approximate weight of any given number of pints of water. 
How could you read the weight of quarts or gallons? Use 
•i- inch for 10 units on both scales, in plotting. 

10. Find the equations of the straight lines joining the 
following pairs of points, timing yourself : 

a. (3, 5) to (- 2, 7). e. (0, 8) to (0, 5). 

b. (3, 5) to (2, -7). f. (1, - 3) to (- 1, - 5). 

c. (0, 8) to (7, 0). g. (1, - 3) to (1, 6). 

d. (0, 8) to (7, - 6). h. (-1, -3) to (-3, -5). 



104 



UNIFIED MATHEMATICS 



i. (-1,-3) to (3,3). 

j. (8, - 3) to (- 3, 2). 

Jc, (-3,5) to (7,0). 

I (-3,5) to (-7, -2). 



m. (-3,5) to (7, -2). 
7i. (2,2) to (-2, -2). 
o. (100, 60) to (0, 0). 



11. Find the equations of the following lines : 



a. of slope 3, ^-intercept 5. 

b. of slope 3, x-intercept 5. 

c. of slope — 2, ^/-intercept 0. 



d. of slope —1, through (2, 2), 

e. of slope + 4, through (2, 2), 
/. having intercepts of 7 and 

5 on the x- and ?/-axes 
respectively. 



4. Distance between two points : d = V \x 2 — orj) 2 + (y 2 — y^ 2 . 
d* = P X P 2 * = P^ + 3/ 2 p 2 2. 

Plilf 2 = #2 — X l} ail< i ^2 A = Vl— V\i 



Since 



d=^/(x,-x i y-[-(y 2 -y i y. 

Whatever the positions of P l and P 2 , parallels to the x- and 
2/-axes through Pi and P 2 form a rectangle (a straight line if 

x x = a? 2 or 2/ 2 = y 2 ) whose 
sides are in absolute value 
| x 2 - ^ | ' and | y 2 — y x | ; 
the bars indicate that 
only the numerical value 
is considered. As a posi- 
tive distance P\M 2i if 
x l > x 2 , would have to be 
written x x — x 2 
or —(x 2 — x 1 ). 
But since the numerical 
value of the expression 
(x 2 — Xi) 2 is the same as 
x 2 ) 2 we may use in every case (x 2 — x x ) for 




Distance between two points 

d 2 =(x 2 -x 1 y + (y 2 -y 1 y. 



the value of (x x 

P X M 2 in the above expression for d wherein only the square of 

P\M 2 enters. 

d = P X P 2 = Vfe-^a^ + te-*/!) 2 . 



STRAIGHT LINE AND TWO-POINT FORMULAS 105 



The distance from any point (x 1} y^ to any point (x 2 , y 2 ) is 
given by this formula ; this distance is taken in general as a 
positive quantity. 

This formula may be used to derive the equation of the 
straight line joining Pi(#i, y Y ) to P 2 (x 2 , y 2 ) for any point P(x, y) 
on the line is such that PP 1 + PiP 2 = PP 2 ; and for no point 
not on the line is this relation true. 

*i*2 + k 2 x l , _ k x y 2 + ^iVi 
k x + k 2 ki + k 2 

"^ i i i ^" i i i ^" 



5. Point of division formula : x 3 



A A A A A A A A A 

PiP 2 + AA = AA- 
For any three points P lt P 2 , and P 3 on a directed line we 
have P^ + ,P 2 P3 = PiP 3 ; if P lies between P l and P 3 , all 
three segments have the same algebraic sign but otherwise 
positive and negative segments are involved. 

OPi + P : P 2 = OP 2 is then, similarly, the fundamental rela- 
tion true for any three points on a directed line, whence 

P X P 2 = OP 2 - OP 1 = x 2 - x,. 
In words the distance on the a>axis (or any other line parallel 
to the a>axis) from any point whose abscissa is x x to any point 



whose abscissa is x 2 , is 
respect to points on the 
y-Sixis, or two points on a 
line parallel to the ?/-axis, 
the distance from the 
point whose ordinate is y x 
to the point whose ordi- 
nate is y 2 is y 2 — y lm 

To find the coordinates 
of the point P 3 which 
divides the line joining 
P]P 2 into two segments 
which bear to each other 



given by x 2 — x x . Similarly with 




the ratio 



fei 



note that 



Point of division formula 
A X A Z _ B X B Z _ 



PiPs = h 
P*Pz k 2 



A%A 2 B3B2 



106 UNIFIED MATHEMATICS 

P P k 

1 3 = ^» By drawing lines through P 1} P 2 , and P 3 parallel 

ijl 2 A/O 

to the axes, similar triangles are formed, or the proposition of 
plane geometry that a series of parallels cut off on trans- 
versals proportional parts may be directly used. 

PiM 3 _ A X A 3 __ x 3 — x t _ k x 
M S M 2 A S A ? x 9 — x* k 
Whence 

, K 

M. + Vi_ ** _*i + rx 2 wherein r=^l. 



x» = 



Similarly, ?A = h- whence y% ~ Vl = &. 



Vs = 



flJ 2 #3 


*i + rx 2 


1 + r 


2/.s - ?/l . 


2/2-2/3 


l/i 4- >#2 



*1 + ^ 1 + ^1 f r 



Vi H — 2/ 9 
**±_** = ' *2_" = * + W , wherein r = *t 

*1 + *2 1 + *J 1 + r fc 2 

A?2 

If P 3 (a? 3 , ?/ 3 ) divides the line PiP 2 externally in the ratio 

k 

— , or r, the segments must be regarded as of opposite signs and 

/to 

k 

consequently, the ratio — , or r, is negative. Either k x or k 2 can 

/Co 

be regarded as negative ; shifting the sign from k 2 to k x is 
equivalent to changing the sign of the numerator and denom- 
inator in the value of x 3 and y 3 , no change is necessary in our 
above derivation of the values of x z and y 3 . 

By eliminating k x and k 2 between the two equations, 

_ k Y x 2 -f- k 2 x 1 
ki -f & 2 

y= ^l2/2 + ^22/l ? 

/C]_ "" f~ lC 2 

the equation of the straight line joining P x and P 2 is ob- 
tained. 



STRAIGHT LINE AND TWO-POINT FORMULAS 107 

Mid-Point : 

Place ki = k 2 , or place r = 1, 

v = y±±y2 
2 ' 

This mid-point formula is of such frequent use that it 
should be separately memorized ; the truth of it is obvious 
from the figure. 

PROBLEMS 

1. Find the mid-point, the points of trisection, and the point 
dividing the segment externally in the ratio, — 2 : 5, in each of 
the following line segments : 

a. (-3,4) to (6,7). b. (3, -2) to (-5,4). c. (0, 0)to (9, 12). 
Locate the points in each of these line segments both graph- 
ically and analytically ; find the length of each segment. 

2. Find the length and slope of the line joining (8, — 3) to 
(- 4, 2). 

3. Plot the graph oi5y + 2x-5 = 0. 

4. Plotp = 5Z + 50. 

5. Find the equation of the straight line joining A(— 2, 5) 
to B(3, 7). Find slope of this line. Find length of AB. 
Find the point of trisection nearest A. Find a point on BA 
extended that divides the segment BA externally in the ratio 
1 : 2. 

6. Given that the velocity of sound at 0° C. is 1090 feet 
per second, and at 30° C. is 1150 feet per second, find the 
velocity at 20° C., assuming that the relation is linear ; the 
point dividing the line joining (0, 1090) and (30, 1150) in 
the ratio 2 : 1 will give the velocity as the ordinate. At what 
temperature will the velocity be 1100 ft. per second ? What 
are the velocity and temperature at the middle point of the 
range given? 



108 UNIFIED MATHEMATICS 

7. The resistance of wire increases uniformly with the 
temperature, r = r (l + at), the rate of increase depending 
upon the material of the wire ; r is the resistance at 0° C. and 
a is a constant. If a given piece of wire has a resistance of 
200 ohms at 10° C. and of 208.4 at 30° C., find the resistance at 
the middle point [of (10, 200) and (30, 208.4)]. Find the equa- 
tion for r in terms of t. Find the value of r when t = ; 
interpret ; find the value of t when r = 0. The theory is that 
at a temperature of absolute zero (—273° C. or thereabouts) 
the resistance would be zero. Ans. r = 195.8 + .42 t. 

8. The resistance of copper wire of fixed diameter varies 
with the length. If the resistance of 1450 feet of a given 
spool is 184 ohms, and the resistance of feet is ohms, find 
the equation for r in terms of I. Plot (0, 0) and (1450, 184). 
What would be the resistance of 5280 feet of this wire ? 

9. Between (—1, 5) and (8, 37) insert 9 points dividing 
the line into ten equal parts, using the formulas 

^ _ K l X 2 + k 2 X x an( J y __ "q?/2 + &2?7l . 

rearranged as follows : 

^ _ k 2 x^ 4~ k l x 1 -f- KiX 2 — ^i^i __ x , fei ( x _ x \ 

Ki -\-k 2 K\ + k 2 

and similarly k, , , 

y = yi + , , , 0/2 -yO- 

Note that k L + k 2 is constant, 10, and k ± changes for the nine 
points from 1 to 9. Use this method in problems 10, 11, and 
14-17 below. 

10. Between (21, .3584) and (22, .3746) insert 5 values di- 
viding the interval into 6 equal parts. 

11. Between (10, .3611) and (20, .3638) insert 9 values di- 
viding the line joining these points into ten equal parts. 

12. Find the point P 3 dividing the line joining Pi(— 1, 5) 
to P 2 (8, 37) externally in the ratio 1 to 7 ; externally in the 



STRAIGHT LINE AND TWO-POINT FORMULAS 109 

ratio Jj ; externally in the ratio ±. Note that either \ or k 2 
must be made negative, or r taken as negative. 

13. Find the point dividing (21, .3584) to (22, .3746) ex- 
ternally in the ratio i, J, -|. 

14. log 1 = 0; log 2 = .301 ; find 9 values dividing (1, 0) 
and (2, .301) into ten equal parts. Compare with the loga- 
rithms of 1.1, 1.2, 1.3, 1.4, ••• 1.9. See problem 9 above. 

15. log 200 = 2.3010 ; log 210 = 2.3222 ; insert 9 values be- 
tween (200, 2.3010) and (210, 2.3222), comparing with the 
logarithms of 201, 202, ••• 209. 

16. log 200 = 2.3010 and log 201 = 2.3032 ; insert 9 values 
between (200, 2.3010) and (201, 2.3032) and interpret. 

17. Given y = 32 x — 17 ; find the corresponding values of 
y when x = 10 and x = 20. Find the points dividing this 
line in the ratio i \, 4, f, \. What points of division are 
obtained ? 

18. Given Pi(— 1, 5) and P 2 (8, 37) ; on the line joining 
these two points find the point whose abscissa is 3, without 
finding the equation of the line. Find the point whose ab- 
scissa is 7. Find the point whose ordinate is 16. Find the 
point whose ordinate is 0. Find the point whose abscissa 
is +14. 

19. Eliminate r between the two equations x=~ — — — - 

1 + r 

and y = ^^ — — • These two equations constitute what are 

known as parametric forms of the equation of the straight 
line joining (—1, 5) to (8, 37). 

20. W^rite the equations of the line joining (3, — 2) to 
(15, 28) in parametric form. Find 6 points on this line. 

21. Prove analytically that: a. The medians of any tri- 
angle meet in a point, trisecting each median, b. The diago- 
nals of any parallelogram bisect each other. 



CHAPTER VII 



TRIGONOMETRIC FUNCTIONS 



1. Angles and angular measurement. — The angle made by 
any line OP with the horizontal line OX is regarded as 
generated by a moving line, an arm or ray, starting from the 
position OX and turning about the point as a pivot, moving 

always in the same plane. This 
moving ray if rotated in the sense 
contrary to that in which the 
hands of a clock move, counter- 
clockwise, is regarded as gener- 
ating a positive angle ; clockwise 
rotation generates a negative 
angle. A natural^ unit of angular 
magnitude is the complete rota- 
tion which brings the moving arm 
back to its original position. This 
measure is used in giving the 
speed of rotation, e.g. the angular 
speed of rotating shafts and wheels is measured in revolutions 
per minute or per second. The angle generated when the 
moving ray is in the same straight line with its original posi- 
tion, but extending in the opposite direction, is called a straight 
angle ; half of this angle is the right angle, which was probably 
the earliest measure of angles used. Thus our terms acute and 
obtuse relate to the right angle. If the angle is conceived as 
given by the relative position of two lines non-directed, it is 
evident that only angles less than a straight angle would be 
discussed. 

110 




Angle generated by rotation 



TRIGONOMETRIC FUNCTIONS 111 

In the ancient development of geometry the right angle, so 
necessary in building, was fundamental ; in Greece up to 
about 150 b.c. the right angle was used as the unit of measure. 
The artificial division of the complete rotation into 360 equal 
angular units called degrees is due to the Babylonians, who 
made this subdivision as early as 1000 b.c The Babylonians 
used 60 as a unit of higher order much as we use ten, and it is 
probable that they divided another natural angular unit, one 
sixth of a perigon, given by the easy construction of a regular 
hexagon, into 60 equal parts called degrees ; each degree was 
divided by them into 60 minutes (partes minutiae primae, in 
Latin, whence " minutes ") and the minute into 60 seconds 
(partes minutiae secundae). 

Another natural system of measuring angles is of funda- 
mental importance in mathematical work. This is the circular 
system, in which the unit angle, called a radian, is the angle 
measured at the center of a circle by an arc whose length 
is the radius. The radius can be laid off on the circle 
2w, 64 or 6.2832, times, and since equal angles at the center 
are intercepted by equal arcs on the circumference, this angle 
can be placed 2-n- times around the center, or approximately 
6£ times in a complete revolution. Just as 1° is used for 1 
degree, so l r is used for 1 radian, and similarly for other 
numerical values ; when no angle sign is used radians are 
understood. 

2 7r r = 360°. - = 60°. 

3 

7r r = 180°. —=30°. 

6 

-=90°. -=45°. 

2 4 

The student should accustom himself to expressing angles 

in radians, particularly the angles of 30°, 45°, 60°, and 90°, and 

those which depend directly upon them. 

Thus 160° = — : 135° = — — ; or, with many writers, simply — desig- 
6 ' 4 4 

nates 135° in radians. 



112 



UNIFIED MATHEMATICS 



A natural system of measurement of angular magnitude 

The arc AB equals the radius OA ; arc A'E'= OA' ; arc A"E" = OA". 




Radian system of measuring angles 



In these circles it is true that 



AB A'B' A"B' 



It is 



OB OB' OB" 

evident that if the circumference of one of these circles is 
divided into any number of equal parts, then lines from to 
these points of division will divide, when extended, the other 
circles into the same number of equal parts. The angle at 
the center is measured, we may say, by the intercepted arc ; 

or $ = — , wherein a stands for the length of the arc and r for 
r 



TRIGONOMETRIC FUNCTIONS 113 

the radius. The length of the arc is given by the formula, 
a a rO, when 6 is measured in radians ; the area A of the cor- 
responding sector of angle is A=^ f 2 0. Since 3.141593 r = 180°, 

j. = 18 ° 3 = 57.29578°, 1° = .0174533 radian. 
3.141593 

2. Quadrants. — 

The two lines OX and T divide the plane into four quad- 
rants, numbered as indicated on the preceding diagram, I to 
IV ; we will commonly designate a quadrant by its numeral. 
In trigonometric work we conceive angles as placed with the 
vertex at 0, one arm falling upon the OX axis to the right, and 
the other arm falling in one of the four quadrants, or upon one 
of the axes. We think of the angle, in effect, as generated by 
an arm rotating about from the initial position OX. Under 
this assumption it is evident that the terminal arm of an angle 
may fall in any quadrant either by a positive rotation or by a 
corresponding negative rotation, the difference between the 
two angles being 360°. Rotations of greater than one revolu- 
tion reproduce in order the positions on the diagram produced 
by rotations of less than one revolution, e.g. angles of 30°, 
-330°, +390°,+ 750°, - 690°, and in general terms, n x 360° 
+ 30° where n is any integer, are represented by the same 
figure. In radians we may say that a r and (2 7ur + a r ) are 
represented by the same diagram for all integral values of n. 

PROBLEMS 

1. Using 34- for tt, compute the value of l r in degrees. 
What is the percentage error ? 

2. Using 3^ for tt, compute the value of 1° in radians. 
Percentage error ? 

_r Ft r 

3. (rive the value in degrees of i revolution; 7r r ; — ; : 

6 6 

IK r r Q r 

1 straight angle ; f of one right angle ; 3 r ; - tt* ; — ■; : ~-; -~ 

5, £ O 4 4: 

7T r 



114 UNIFIED MATHEMATICS 

4. Give the value in radians of 1 revolution ; 180° ; 45° ; 
135° ; 60° ; 120° ; 225° ; 3 right angles ; 390° ; 765°. 

5. What is the percentage error in using 57.3° as the value 
of 1 radian ? 

6. What error in seconds is introduced by using 57.3* for 
1 radian in finding the value of 3 radians ? 3 r = 171.9° ; an 
error of 1 % would be approximately 1.7°. 

7. A bicycle rider pedals at the rate of 20 miles per hour ; 
how many revolutions does the rear wheel, diameter 28 inches, 
make per minute ? The rear sprocket wheel, diameter 4 inches, 
makes the same number of revolutions as the rear wheel ; how 
many revolutions does the front sprocket wheel, diameter 
10 inches, make ? Changing gear shifts the chain to a smaller 
rear sprocket ; what speed will be attained at the same rate of 
pedaling by shifting to a 3-inch rear sprocket ? 

8. Place the following angles in their proper quadrants : 

150°, 240°, 760°, -840°, 1^, —, - — ir\ Give the cor- 
' 3 4 3 

responding positive angles less than 2 if. 

9. In the circle of radius 10 what is the length of the arc 
of an angle at the center of 60° ? What is' the difference 
between an arc of 60° and an angle of 60°? What is the 

length of the arc of 30°, 45°, -, — ? 

6 6 

10. What is the angle at the center in radians and degrees, 
in a circle of radius 100, subtended by an arc of length 100 ? 
50 ? 30 ? 100 7r ? Find the areas of the corresponding sectors 
of the circle. 

11. In the artillery service angles are measured in "mils"j 
a " mil " is defined as 6 * 00 of a complete revolution. Com- 
pute the value in radians of one mil. 

12. On the " mariner's compass " the complete revolution is 
divided into 32 parts, called " points " of the compass ; com- 
pare the " points," with degrees, " mils," and radians. 



TRIGONOMETRIC FUNCTIONS 115 

13. Compute the value of the "mil" in minutes and give 
approximate formulas for converting " mils " into minutes 
and conversely. 

14. At what rate per second in degrees, and in radians, do 
the hands of a clock turn ? 

15. A grindstone of diameter 18 inches is turning 246 times 
per minute. Compute the linear velocity of a point on the 
rim. 

16. In grinding certain tools the linear velocity of the 
grinding surface should not exceed 6000 feet per second. 
Find the maximum number of revolutions per second of a 
10-inch (diameter) emery wheel and of a 5-inch wheel. 

17. Find the angular velocity in revolutions and in radians 
of an Ohio grindstone, 2 feet in diameter, which should have a 
circumferential speed of 2500 feet per minute. 

18. The path of the earth is approximately a circle with 
radius 93,000,000 miles ; find the distance traveled in 1 day. 
What percentage of error would be introduced by using 365 
instead of 365^ days ? Show that the fact that we give r as 
93,000,000 implies that the position of the point on the earth 
would not affect our computation. 

3. Polar coordinates and angular variables. — Any point P 
in the plane may be located by giving its distance from a 
fixed point O, called the pole, and the angle which a line from 
the pole to the point P makes with a fixed line OR, called the 
polar axis. In general terms the polar coordinates of any 
point, of a variable point, are designated by r and 0, radius 
vector and vectorial angle. (See p. 116.) 

r will be assumed to be a positive quantity, and may 
be assumed as the angle generated by the rotation of the 
vector OP from an initial position on OR. A negative angle 
is generated with the polar axis by a line which turns from the 
polar axis, about O, in the clockwise direction. Thus the 
Z-ROP is taken as + 30° ; this same figure may also be con- 



116 



UNIFIED MATHEMATICS 



i« o ic o in o inoioo' ° o o o op o ~ o o " • 

«S< tj< CO CO NNHHOOiOOiOOOOinO uC O uO O lO 



150 
155 
160' 
165' 
170 ! 
175' 
180' 
185' 
190 
195' 
200 
205' 
210' 












30 
25° 

20° 
15° 
10' 
5' 

355° 
350° 
345' 
340' 

335° 



lO O lO O ^ O >-t OOOlCOoOlCO i.t O lO O tO O lO 

^H O) CM CO CO — -r <_- i ~ ' £ ■- I ■» l^ X 3C » OS O O t-i r-i CN) CM 

CM CM C^ (N C^ CIMMMM^M^^MMMW CO CO CO CO CO 

Polar coordinate paper 

Location f (1 ?' 3 ° 0) ' (10 ' 15 ° 0) ' (10 ' " 15 ° 0) ' a " d (1 °' ~ 3 ° 0) ' 
of {(10, g, (10, «*), (10, =|=), and (lO, ^). 

ceived as representing — 330°. Angles which differ by multi- 
ples of 360°, generated by lines rotating from an initial posi- 
tion upon the polar axis, are represented by the same diagram ; 
two such angles are commonly called " congruent " angles. 
Each rotation of 360° brings a line back to its starting place. 



PROBLEMS 



1. Locate the points (3, 30°), (6, 90°), (4, 45°), (8, 135°), 
(3, 270°), (6, - 90°), (5, 180°), and (2, 390°). 

2. Locate the points (5, ~\ (4, -^-V (6, 0), (3, — — 



(V2, 7r r ), and (3,3^). 

3. What is common to all points on OR ? 
• 4. What curve is represented by r = 10 ? 



5. What curve is represented by = 30° or = — ? 



TRIGONOMETRIC FUNCTIONS 



117 



4. Trigonometric functions — sine and cosine. — Assume an 
ic-axis to coincide with the polar axis, and a ?/-axis to be drawn 
perpendicular to 
the polar axis at 
the pole. When 
6 is any fixed 
angle, the coordi- 
nates (x, y) in 
rectangular coor- 
dinates and (r, 0) 
in polar coordi- 
nates, of points 
upon the ray 
making the angle 
6 with OX, are 
connected by the 
following relations 

y -l = -ll = 





i 


l_ X 










































f%.*** - ' 




r^«K X 


t 


-AS? U 




*^i i?/ 




1 ^3 


"s%- 9 ^- 






: t: ~r 








? /- -t 


) !S^ 1 


V \ 2 


1 ^ 


_l_ 


-A- ** 1^/, 1 


1 X 


^*\T i \ 




J2. ] X i x I 




*'^ %- Li 


: : c 


tt. ■ p! ■ |_ ' 1 








Oz "ir*-- -*M^ li , 






it --. 
















I i 




i i ii <* 





= ZZOP; sin5 



Ml = Ml 



K-; COS* 

r 



and 



and ^ 



■*'2 
**2 

^ + 2/2 



= -, for points upon the ray, 



for any point (x, y) in the plane. 



We may say that - is a constant for any given angle 0; 

this constant changes as changes. It is evidently a function 
of 0. Since r remains positive, this function is positive for 
all angles represented in the upper quadrants ; negative for 
angles in quadrants III and IV. This constant is \ f or 6 = 30° 
V2 
2 
6 = 90°, .866 for = 120°, .707 for = 135°, \ for = 150°, and 
for 6= 180°, all by elementary geometry. When is an angle 
which lies in quadrant III or IV, i.e. values of 6 between -f- 180° 



or 



?> 



or .707 for 6 = 45 c 



i V3 or .866 for = 60 c 
135°, i for 6 = 150° 



lfor 



118 



UNIFIED MATHEMATICS 



and + 360°, this function of becomes negative. This function 
of 6 is called the sine of 0, or sin.0. 

sin0 = ^« 



■n 



*ifi-- 



^\ 



m 



% i 






E 



:T: 



sin* =^- 
r 



cos0 



Polar coordinates, r and 
Rectangular coordinates, x and y 



Similarly the ratio 



x 
r 

is a constant, whose 
value depends entirelv 
upon the position of the 
moving ray; this func- 
tion of we define as 
cosine 0. 



COS0 = 



sin0 = 



The consideration of the changes in value of these functions of 
6, sin 6, and cos 0, as changes, is facilitated by thinking of the 
moving ray as fixed in length. 

For positive values of 6 less than 90°, in I or, in symbolic 
language, < < 90°, it is evident that the complementary 
angle to any angle gives a tri- 
angle similar to the triangle in- 
volving 0. In this second tri- 
angle the ordinate and abscissa 
correspond respectively to the ab- 
scissa and ordinate in the original 



Now 



triangle, whence ^- = -< 
r r 

t- = sin (90° - 0), and * = cos 6 ; 
r r 

hence cos = sin (90° — 0), or, in 
words, the cosine of any angle 
6 (0 < < 90°) is the sine of the 
complement of 0. This explains 




Complementary angles, 
+ 0i = 90° 



the name, cosine 0, which is simply the " complement's sine." 
Further, — = - , whence cos (90° — 6) = sin 6. 



TRIGONOMETRIC FUNCTIONS 119 

Either one of the triangles may be regarded as the origi- 
nal, the complementary angle will be found in the other; 
the demonstration, as given, applies in either case. The 
above figure serves, then, to demonstrate the two formulas, 
sin (90° — 0) = cos and cos (90° — $) = sin 0, for any positive 
acute angle 0. Later these formulas will be shown to hold for 
all angles 0, without restriction as to magnitude or sign. 

The formula cos (90° — 6) = sin may be derived from 
sin (90° - 0) = cos by substituting for the value 90° - 0', and 
finally replacing & by 0. Since 6 may vary from to 90°, 
90° — 6 varies between the same limits. 

sin (90° -6)= cos 9, 
cos (90° — 6)= sin6. 

5. Historical note. — The function sin is Hindu in its 
origin, dating back probably to the fourth century a.d. The 
Hindus called the sine " ardha-jiva," meaning half-chord. In 
the eighth century a.d. the Arabs becoming familiar with Hindu 
astronomy and trigonometry, as used in astronomical work, 
transliterated the word " jiva " or " jiba " into " geib " ; the 
word in Arabic means curve and in the twelfth century Euro- 
pean translators into Latin of Arabic works of science trans- 
lated this word as " sinus." Into English the word comes by 
transliteration again, the sound and. not the sense being pre- 
served. 

Plane trigonometry is possible using the chords instead of 
the half-chords ; this system was developed by the Greeks, but 
it leads to much more complicated formulas and methods. 



6. Tangent and the reciprocal functions. — The quotient 



sin 6 

cos 6 

varies as varies ; this is then a function of 0. This function 
is called the tangent. By definition, 



ton 6 = ^ MY 

cos 6 V x ) 



120 UNIFIED MATHEMATICS 

The reciprocals of sin 6, cos 6, and tan 6 are also functions of 
6 ; to these the names cosecant 6, secant 6, and cotangent 6 
have been given. The six fundamental definitions follow : 



sin 6 = -, cosecant 6, or esc = 



r 



sin 6 



x 1 

cos 8 — -, secant 6, or sec 9 = 



r 



cos 8' 



tan 8 = = -, cotangent 6, or cot 8 = 



cos 6 x tan 8 

PROBLEMS 

1. Given sin 6 = .29, find cos 6 using the formula 
sin 2 6 +cos 2 = 1. The negative value has a meaning. 

2. In what quadrants is sin0 positive? in what quadrants 
is cos 6 positive ? 

3. Given sin 6 = .29, in what quadrants may lie ? 

4. In what quadrants is tan positive ? 

5. As a rotating arm of length 10, moving about from 
OX, turns through 90°, discuss the changes in value of the y of 
the end of the moving arm ; consider r as 10 and discuss the 
change in value of sin as the angle generated' increases from 
0° to 90° to 180°. What change in sin 6 as 6 increases beyond 
180°? 

6. Discuss similarly the changes in values of cos as 6 
varies from 0° to 90° ; from 90° to 180°. 

7. Discuss the possible values of tan 0. Take x = 1, ^, .1, 
.01, .001 and compute y in a circle of radius 10. Discuss the 
values of tan 0. When x = .000001, y = 9.99999999999995 
what is the approximate value of tan 6 ? 

8. Given tan 6=3, find sec from the formula 

sec 2 (9 = 1 + tan 2 0. 
Compute both the positive and the negative values of cos 6, 

9. Express in terms of the sine of the complementary 
angle : cos 48°, cos 84°, cos 56°, cos 48° 10', cos 90°. 



TRIGONOMETRIC FUNCTIONS 121 

10. Express in terms of the cosine of the complementary 
angle sin 48°, sin 81°, sin 56°, sin 48° 10', sin 90°. 

11. Complete the following table : 



cos 


45° 


= 


.7071 


= 


sin 


45 c 


cos 


46° 


= 


.6947 


= 


sin 


44" 


COS 


47° 


— 


.6820 


— 


sin 





cos 48° = .6691 = sin 
cos 49° = .6561 = sin 
cos 50° = .6428 = sin 
Reverse the table, beginning sin 40° = 

sin 41° = 

12. Complete the following table : 

sin 35° = .5736 = cos 
sin 35° 10' = .5760 = cos 
sin 35° 20' = .5783 = cos 
sin 35° 30' = .5807 = cos 
sin 35° 40 ' = .5831 = cos 
sin 35° 50' = .5854 = cos 
sin 36° = .5878 = cos 

Notice that the sines of 35° + some minutes are cosines of 
angles 54° + some minutes ; the cosines of 35° + minutes are 
sines of the complements, 54° -f- minutes. In our tables you 
have written at the left of the table 35° and 54° at the right ; 
sin at the top and cos at the bottom. 

35° 



r 


sm 


COS 







.5736 


.8192 


60 


10 


.5760 


.8175 


50 


20 


.5783 


.8158 


40 


30 


.5807 


.8141 


30 


40 


.5831 


.8124 


20 


50 


.5854 


.8107 


10 


60 


.5878 


.8090 





i 


cos 


sin 


i 



54 c 



122 



UNIFIED MATHEMATICS 



7. Fundamental formulas. — Since x 2 + y 2 = r 2 , for any point 
on this circle of radius r, 



© 



+W-i. 




Note that although x or y or both may be negative, the rela- 
tion continues to hold since (— x) 2 = x 2 , and (— y) 2 = y 2 . 
Whence, by substitution, cos 2 + sin 2 = 1, for all values of 6. 

By division by cos 2 0, 

sin 2 6 1 



1 + 



cos 2 <9 cos 2 0' 



or 1 -f tan 2 6 = sec 2 0, for all values of 0. 
Similarly, 1 + cot 2 6 = esc 2 6. 

sin 2 6 + cos 2 9 = 1. 
1 + tan 2 6 = sec 2 8. 



TRIGONOMETRIC FUNCTIONS 



123 



These formulas are of fundamental importance. They 
should be memorized. 

8. Functions of 0°, 30°, 45°, 60°, and related angles. — By 

plane geometry the values of these functions can be precisely 
determined for the angles which can be geometrically con- 
structed with ruler and compass. The 
most important of these angles are 30°, 
45°, 60°, and 72° ; the values are evident. 
for and 90° (as limits). 

1 = V2 
V2 2 



sin 45 c 



'07. 



cos 45 = — - == — — = 
V2 2 

tan 45° = 1 = cot 45°. 



'07. 

























































(y 








l- 






















































i 









Functions of 45° 

One half a unit 
square . 



















































30 : 
















2, 










\ 2 
































, 


































1 






1 







sin 60° = 



tan 60 c 



tan 30° = 





^ 


~3.'^4, 


*> JQ-1 


s'W 


Zk 




2 S 


^\ 





V3 
2 
= cos30 c 
cos 60° = \ 

= sin30 c 

V3 

cot 30 c 

3_ 

V3 

= cot 60 c 

1_ = V3 
2 -.-.. v _ 3 

These diagrams should be memorized as half of a unit 
square for 45°, and half of an equilateral triangle placed ver- 
tically for the functions of 60° and directly related angles, 
and the same placed horizontally for the functions of 30° and 
related angles (-30°, 150°, 210°). 

sin0° = 0. sin 90° = 1. 

cos 0° = 1. cos 90° = 0. 

tan0° = 0. tan 90° = oo. 



Functions of 60° 
Equilateral triangle. 
V3 



Functions of 30° 

Equilateral triangle. 



= .866. 



= _^ = .577. 



124 



UNIFIED MATHEMATICS 



The meaning of the expression tan 90° = oo (infinity) is that 
as the angle 6 approaches nearer to 90° the tangent becomes 
larger than any quantity we may assign, however large; 
strictly at 90° the tangent function has no meaning, as a divi- 
sion by zero is involved. The expression tan 90° = oo is not, 
then, an equality, like tan 60° = V3. 



36 



1 



I 



m 



Construction of the regular decagon 

OM divides OA in " extreme and mean " *atio. 

Algebraical method by solving, x 2 = 10 (10 — x). 

The method of constructing a decagon combined with the 
solution of a quadratic equation enables us to find the sine of 
18°. The radius of the circle is divided in extreme and mean 
ratio to obtain the side of the inscribed decagon : 10(10— x)=x 2 , 
in a circle of radius 10. Whence, 

x 2 + 10 x - 100 = 0, 

x = -5± VI25 = - 5 ± 11.1803 = - 16.180 or + 6.1803, 
of which we take the positive value. One half of this value 
is the value of y in the triangle of reference for 18° when 
r = 10. Hence the sine of 18° is 
3.090 



10 



= .3090. 



TRIGONOMETRIC FUNCTIONS 



125 



9. The Greek method, using chords. — By the methods of plane 
geometry, using chords instead of half -chords, the sine of half an angle 
and the sine of the sum and the difference of the two given angles can be 
computed. One theorem involved, in addition to the Pythagorean 
theorem, is not given in many geometries. It is called Ptolemy's the- 
orem, as it is fundamental in the method of computing chords de- 
veloped by Ptolemy, a Greek writer of the second century a.d., whose 
text-book on astronomy, the Almagest, continued in active use for fifteen 
hundred years. The theorem is that in an inscribed quadrilateral the 
product of the diagonals is equal to the sum of the products of the oppo- 
site sides. 

From the chord of 60° one can compute the chord of 30° ; thus the 
sine of 15° is obtained. Prom 36° and 30° the sine of 3° can be obtained 
by using half the chord of the difference of two given arcs ; from this the 
sine of li°, f°, |°, T 3_o 9 _3_o 5 _3_o 5 _|_o ? _^o 5 ... can be computed> The 

sine of 1° camiot be obtained by this process, nor can the sine of i° ; 
these are found by other methods, giving approximations as accurate as 
desired for any practical purposes. 

10. Origin of the tangent and cotangent functions. — In the 

study of astronomy the angle of inclination to the horizon of 




Arabic shadow function 

The shadow varies as the cotangent of the angle of inclination of the sun. 



the sun and of other heavenly bodies is important. The ratio 
of the length of the shadow to the length of the vertical object 



126 



UNIFIED MATHEMATICS 



casting the shadow gives the cotangent of the angle of in« 
clination of the sun. This function of the angle appeared 
before the tangent function in the works of the Arabic as- 
tronomer, Al-Battani, of the tenth century a.d., and it wa? 
called the shadow and later, right shadow or second shadow. 
The tangent function, being the ratio of the length of the 
shadow cast on a vertical wall to the length of a stick placed 
horizontally out from the wall, was called later the first 
shadow. The Arabs took the length of the stick as 12. 




Variation of sine and cosine as 6 varies. 



11. Variation. — As varies the trigonometric functions 
also vary ; it is desirable to fix in mind the changes of the 
three principal functions, viz. sin 0, cos 0, and tan 0, as 
changes by rotation of the moving arm. 

Taking r = 10, it is an easy matter to follow on the graph 



TRIGONOMETRIC FUNCTIONS 127 

the changes in the x and y of the end of the moving ray. As 
the moving ray starts from OX, an angle of 0°, the y or ordi- 
nate is zero. So we have that the sine, ^, begins at zero for 

r 

0° ; as increases the y increases, reaching a maximum of 10 
when is 90° and the maximum value then of sin is \% or 1. 
As increases beyond 90°, the ordinate begins to decrease, 
arriving finally at when the moving arm is on OX'. For 
angles greater than 180° up to 270° the ordinate decreases, 
finally reaching a minimum or lowest value of — 10 ; the cor- 
responding minimum of sin is — 1 ; from 270° on to 360°, 
completing a revolution, the sine increases from — 1 up to 0. 

For angles greater than 360°, or for negative angles, the 
moving ray would move through no new positions ; for any 
such angle the trigonometric functions are equal to the func- 
tions of the corresponding positive angle having the same 
position. 

The limits + 1 and — 1 of the sine function and cosine 
function are evident, of course, in the figure. In any position 
of the moving ray x and y are the sides of a right triangle of 
which r is the hypotenuse, except that on the axes x or y 

II x 

equals r ; hence the quotients - and - are either numerically 

less than 1 or at most equal to 1. 

Note particularly on the diagram the sines of 30°, 45°, and 

K 7 "1 o 7 

60°, as — , approximately -^-, and -^-: the values, .500, 0.707, 
10 F J 10 10 

and 0.866 may well be memorized. On the diagram it is a 

simple matter to read the sines of 10°, 20°, 30°, 40°, 50°, 60°, 

70°, 80°, and 90° as the corresponding ordinates divided by 10, 

correct to two decimal places. The cosines of these angles are 

read as the corresponding abscissas divided by 10. 

The tangent as - is not in a form to give the numerical value 
x 

without computation ; however, by drawing the tangent line 

to the circle at A and producing r to cut the tangent line at 



128 



UNIFIED MATHEMATICS 



T, you have — — = ^ ; whence 



tan a, and so the value of 



^=£; whence *?: 
OA x 10 

the tangent of the angle can be read as the ordinate at A di- 
vided by 10. 



X 



5 



-M: 



-Azzl 



tan 30° = 



V3 10 



tan 45 c 



"io" 1 ' 

V3 17.3 



tan 60° = — = 



= 1.73, 



The tangent read as a length 
AT =10 tan a. 



1 10 

When the angle increases 
beyond 90° the position of the 
terminal arm fixes the sign 
of each function ; the sine is 
positive when the arm is in 
the upper quadrants, I and 
II, and negative in the lower, 
and the cosine positive to the 
right, I and IV, and negative in II and III. The tangent is 
positive in I and III, and negative in II and IV ; when posi- 
tive the corresponding vertical lengths are cut off above A on 
the tangent at A, and when negative, in II and IV, the corre- 
sponding vertical lengths are cut off below A on the tangent. 

If the radius is taken as unity, the ordinate, abscissa, and 
tangent length represent numerically and in algebraic sign the 
sine, cosine, and tangent. values of the corresponding angle. 
However it is usually more convenient to take a radius of 10, 
25, 50, or 100 and to interpret the trigonometric functions as 
ratios, as indeed they are. 



12. Related angles. — From our definitions it is evident that 
sin has the same value for two angles, symmetrically placed 
with reference to the y-axis, 9 and 180° — 6 ; cos 6 has the 
same value for two angles symmetrically placed with respect 
to the a>axis, and — 6, or $ and 360° — j tan has the same 



TRIGONOMETRIC FUNCTIONS 



129 



value for two angles which differ by 180°, 6 and 180° + 0. All 
functions are the same for angles which differ by 360°, or by 
any integral (positive or negative) multiple of 360°, for the 
terminal arms of such angles will coincide when the angles are 
placed in position to determine the trigonometric functions. 

The trigonometric functions of 360° - 0, 180° - 6, 180° + 0, 
90° + 0, 90° — 0, and — 0, in terms of the functions of are of 
particular importance in later work. In the figure the vectors 



m 

HE 



g^? 



wm 



2 






3 



Pi 



Q, - 6, 180° - 6, 180° + B. Related angles 

Eead the corresponding functions on the diagram. 

OP v OP 2 , OP 3 , and OP± are the terminal arms of related 
angles in quadrants I, II, III, and IV. The vector OP x de- 
termines, we may say, a positive acute angle XOP^, and, 
further, any angles which differ from the positive acute angle 
by any integral multiple of 360° ; OP 2 represents the terminal 
arm of 180°- $, OP 3 of 180°+ 0, and OP 4 of 360°- 6, or of - 0. 
If 6 is the angle represented in quadrant 1, 180° — is the 
angle here represented in II ; and conversely, if is in II, 
180°- $ is in I ; if 6 is the angle in III, 180°- is the angle in 
IV, and conversely. Evidently if in I is 30°, 180° - 6 is 150°, 
represented in II, and if = 150°, 180° - 6 = 30° ; further, if is 
- 330° in I, differing from 30° by - 360°, 180° - $ will be 
180° -(- 330°) or 510° which is in II, 360° + 150°, differing 
from 180°- 30° by 360°. The ordinates in I and II are equal 



130 UNIFIED MATHEMATICS 

and of the same sign, and similarly the ordinates in III and IV 

are algebraically equal ; r is the same in all. Hence for all 

angles $, 

sin (180° - 9) = sin 9. 

The abscissas in I and II are numerically equal but opposite 
in sign, similarly in III and IV ; the vectors are the same and 
positive. 

Hence cos (180° — 6) = — cos 9. 

By definition, tan (180° - 0) = Sm (180 ° ~ ^ , for all angles. 
J J cos (180° -0)' 8 

By substitution, 

tan (180° - 9)= sin ^ ~ f > = ^^ = - tan 9; 

v ; cos (180°- 6) - cos 

this formula also holds for all angles 0, since every formula in- 
volved has been shown to hold for all angles 6. 

If 6 is an angle in I, 180° -f- is in III ; the corresponding 
positions are represented for any such angles by our figure. 
If is represented by the vector in II, 180° + 6 is represented 
by the vector in IV. The ordinate in III equals numerically 
the ordinate in I, but is opposite in sign ; similarly the ab- 
scissas of I and III ; similarly the ordinates and abscissas, 
respectively, of II and IV are equal in value and opposite in 
sign. 

Hence sin (180° + 9)= - sin 9, 

and cos (180° + 9)= — cos 9. 

These equalities hold for all angles 6. 

By definition, tan (180° + fl) = si *( 180 °+ *) =^^ = tan 0, 
J > \ -r J cos(180°+<9) -cos0 

which holds for all angles 0. 
In precisely the same way 

sin (— 9)= — sin 9, 
cos (—9)= cos 9, 
and tan (— 9) = — tan 9, for all values of 6. 



TRIGONOMETRIC FUNCTIONS 



131 




6, 90° + 0. Related ang!es which differ by 90° 
Our second figure can be used to show that 

sin (90° + 0)= cos 0, 
cos (90° + 6)=- sin 0, 

for all values of 6. If is the angle represented in I, 90° + 6 
is the angle here represented in II ; if 9 is in II, 90° + is rep- 
resented in III ; if 6 is in any quadrant, 90°+ is in the quadrant 
following in the counter-clockwise sense. Now the ordinate in 
any quadrant here equals numerically and algebraically the 
preceding abscissa ; thus y 2 = x 1 ; y 3 = x 2 ; y 4 

For any angle 6, sin (90° + 8) = cos 8. 

Similarly cos (90° + 8) = — sin 8, 

tan (90 



Vl = «4- 



e , sin (90° + 



cos 



cos (90° + 6) - sin 6 
= - cot 8. 



132 UNIFIED MATHEMATICS 

The following relations have now been established for all 
angles 6 : 

sin (180° - 6) = sin 0, sin (180° + 8) = - sin 0, 

cos (180° - 0) == - cos 0, cos (180° + 0) = - cos 0, 

tan (180° - 0) = - tan 0, tan (180° + 0) = tan 0, 

sin (— 0)=— sin 0, 
cos ( — 0) = cos 0, 

tan (—0) = — tan 0. 
The formulas, 

sin (90°- 0)=cos0, 

cos (90° -0)- sin 0, 

and tan (90° -0)= cot 0, 

have been established for acute angles. However, the preced- 
ing formulas' for 90° + which we have established for all 
values of 6, positive and negative, can be used to prove that 
these formulas for 90° — hold for all values of 0. 

Thus sin (90° — 0) can be considered as sin (90° +(— 0)), 
and as the formulas for 90° -j- hold for all values of 0, we 
have : 

sin (90° -f-(- 0))= cos (- 0)= cos 0, 

and cos (90°+(- 0)) = - sin (-$) = -(- sin0) = + sin0. 

Hence sin (90° - 0) = cos 6, 

and cos (90° — 6) = sin 6, for all values of 0. 

Further tan (90° - 6) = sin (90 ' ' = °± = ?™l = cot 6 y 

y } cos (90°- 0) sin0 

again for all values of 0. 

The student will do well to remember the diagrams and to 
connect the formulas with these. It is necessary to recollect 
only the representation for an acute angle 6 ; it is more desir- 
able to connect the formulas with the diagrams than merely to 
memorize the formulas. 



TRIGONOMETRIC FUNCTIONS 133 

PROBLEMS 

1. Find the sine, cosine, and tangent of 210°. The triangle 
is the same as that nsed for the f mictions of 30°, but it is 
placed in III. 

2. sin 150° = ; cos 150° = ; tan 150° = 

3. sin 315° = ; cos - 45° = ; tan - 45° = 

4. sin 225° = ; sin 495° = ; tan 750° = 

5. Express the following in terms of functions of positive 
angles less than 45° : 

a. sin 170°= b. sin 130° = 

c. cos 170°= d. cos 130° = 

e. sin 220° = /. tan - 40° = 

6. Express in terms of functions of x : 

a. sin (x — 90°) = 

Hint. — Use first sin ( — 0) = — sin 9. 

b. sin (270° + a) = 
Hint. — Express first as 

(180° + 0) ; i.e. sin (180° + 90° + x) = - sin (90° + x) = •». 

c. cos (z-270°)= d. tan (360° - a;) = 

7. Draw one quadrant of a circle of radius 10 half-inches ; 
construct the angles of 30°, 45°, and 60° and read their values. 
Bisect the angle of 30° and so obtain the values of the func- 
tions of 15°. Make a table of values of sines, cosines, and tan- 
gents, advancing by 15°. Note that the chord of 30° may 
readily be computed ; one half of this chord divided by the 
radius gives the sine of 15°. Find the sine of 1\° similarly. 
From the table of sines of acute angles from to 90° by 
15° intervals, give the sines of the related obtuse angles up 
to 180°. 

8. Given tan 6 = 3, find sec 6 and cos 9 ; what is the signifi- 
cance of the double sign in the answer ? 



134 UNIFIED MATHEMATICS 

9. Express in terms of functions of positive angles less than 
45°: 

a. sin 100°. e. tan 200°. i. tan (-420°). 

b. cos 100°. /. sin 300°. j. sin 750° 50'. 

c. tan 100°. g. cos (-60°). k. cos 1030° 40'. 

d. sin 200°. h. cos (-160°). I. tan 218° 10'. 

13. Angles constructed from given functions. — Given 

sin $ = ± , 

construct both values of 9 (in I and II). The problem is 
in geometrical language to construct a right triangle with the 

hypotenuse and one side given, since sin = ^ • 

r 

Take r as 7 and y as 4 ; since sin 6 = ^, 7 is to be one side of 

r 
our angle ; with 7 as a radius and as center describe a semi- 



::::==========:==:^::g==^==:::^::::::::::: 

::::::::::::::::===i=:»= ]lL . — rf II;__. .l'J 

::::::::::i^S::5t#^^ffi::::::::::: 

=i#lllllM 

::::j::i:::::::::^;::i->-:-:::::::}::t::::::: 



«i 



Graphical solution of sin 6 



circle above the oj-axis ; y = 4 is a line parallel to the axis of x, 
cutting the circle, x 1 -f- y 1 = 49, in two points. Find the inter- 
sections % and a 2 ; geometrically the angle is found. Using 
the Pythagorean theorem x 2 -f 4 2 = 7 2 , and x 2 = 33, x = ± 5.75. 
The positive value of x is to the right and the negative to the 



TRIGONOMETRIC FUNCTIONS 



135 



left ; -to the first corresponds the acute angle 0, and to the 
other with the terminal arm in II. 
In I, sin0 =i = .57. 



In II, sin 



n 9. It) ,x 

cos = = .82. 



cos 6 = 



= - .82. 



tan 



4 
5.75 



.70. 



tan = — - — - 
5.75 

= - .70. 



This problem should also be solved using the formula 

sin 2 + cos 2 = 1. 

Given tan = 1.4, find the other 
functions of and discuss the two 
solutions. 

cos x 



can be in I or III. 

Take y as 1.4, x as 1 (or y as 
14, x = 10, or other values as con- 
venient) ; evidently x as — 1 and 
y as — 1.4 gives in III. 











V 


1 




1 




















-l-'i 




• 




















it 








(i 


,1.4) 














A Jk?i 






















hp 


% 
























T ' l 7 
























jjy 


























/A 
























r> 














-1. 


2-l10- 


i-.li- 


^v 


.2 2 .4 . 


j . 


■ 1 


1 


.2 


r 




^T 


/ 




I4 














1 ' ^"a> 


/ 




!e 

!s 

-ilo- 

1 
















' \AW 




















I V 


































































7pw.4) 


















III 

















Graphical so-ution of tan 6 = 1.4 



Since x 2 + ?/ 2 = ?- 2 , r 2 = 1 + 1.96 = 2.96. 

r = 1.72+ 



T T ■ /) 1-4 

In 1. sin = or 



1.4 



1.4 



three places), 
cos 



1-^ V^96 2.96 



&2 = ' 5S1+ ' 



V2.96 = .81 (or .814 to 



In III, sin = - .814, 
cos $=- .581. 

This problem should be solved also by using the formula 
1 + tan 2 = sec 2 0. 



136 UNIFIED MATHEMATICS 

EXERCISES 

1. Use 10 of the larger units on a sheet of cross-section 
paper and find by construction the sines of the angles 0°, 10°, 
20°, 30°, - 180°. Compare with the table. Note that the 
values for 10° and 170°, 20° and 160°, 30° and 150° ... corre- 
spond. Use the formula cos 6 = sin (90° — 0) to find the 
cosines of '0°, 10°, 20°, — 90°. Note that in the second quadrant 
the cosines become negative. 

2. Construct an angle of 60°, using 10 as side of the equi- 
lateral triangle used. Find cos 60°, sin 60°, tan 60° to 2 places. 

3. Find the sine of 150°, 210°, 330°. 

Use half the equilateral triangle, placed horizontally with vertex of 30° 
angle placed at the origin. 

4. Find the sine and cosine of 120°, 135°, 225°, - 30°. 

5. Find the tangent of 120°, 135°, 225°, - 30°, from the 
data of the preceding problem ; find tan 120°, tan 135°, tan 225°, 
tan (— 30°) from the geometrical figure. 

6. By construction of a square, side. 10 units, find approxi- 
mate values of the functions of 45°. Find the values using the 
Pythagorean theorem. 

-7. Construct an angle of 30°, and find values of the 
functions. 

8. Construct angles of 15° and 7-1- , and find the values of 
the functions. 

9. Given sin 6 = -^, find cos and tan ; indicate both 
solutions. 

10. Given cos = .432, compute sin and tan to 3 places, 
0inl. 

11. Given tan = 4.32, compute sin and cos $ for 6 in III. 

12. Given tan 6 = \ , construct 6 geometrically. 

13. Construct 6 geometrically, given sin = T V 

14. Given sin = — ^, find cos 6, in IV. 

15. Given sin = .43, find cos 0, in III. 



TRIGONOMETRIC FUNCTIONS 137 

16. Given tan = — .43, construct in II, and find values 
of sin 6 and cos from the figure. 

14. The inverse, functions. — If we are given the sine s of an 
angle and desire to speak of the angle we can say " the angle 
whose sine is s " and we can abbreviate this expression in 
writing to arc sin s or to sin -1 s. Similarly the angle whose 
cosine is m is written arc cos m, or cos -1 m. Note that in sin -1 s, 
cos -1 m, and tan -1 k, the — 1 is not at all a negative expo- 
nent ; these expressions for angles are read anti-sine s, anti- 
cosine m, and anti-tangent k, or sometimes, inverse sine s, etc., 
respectively. 

In what follows we shall use mainly the symbols arc sin, 
arc cos, arc tan, arc esc, arc sec, and arc cot, although the 
other symbols are also in common use. Whether the " arc" or 
" — 1 " symbols are used the student is strongly advised to 
read " arc tan t " or " tan -1 1 " always as " the angle whose tan- 
gent is t," and similarly expressions like arc cos x, arc sin -J, and 
sin -1 1\ 

A given angle has only one sine, but a given number is the 
sine of many different angles. A similar remark applies to 
the other five functions. To illustrate : sin 30° is 0.5 and no 
other value. But arc sin 0.5, the angle whose sine is 0.5, may 
be 30° or - 330° or 390° or 750° or 150° or 510° or 870° or any 
angle differing from 30° or 150° by an integral multiple of 360°. 
The sine of any one of these various angles is 0.5 ; 

sin (k 360° + 30°) = .5 and sin (k 360° + 150°) = .5, 
where k is any integer. 

PROBLEMS 

1. Given arc cos i = 0, construct both in the first and in 
the fourth quadrant. Note that the problem is precisely the 
same as though the requirement were to construct when 
given that cos = 1; or to construct arc cos i. 

2. Between what values must k lie to have any solution for 
$ = arc cos k ? 



138 



UNIFIED MATHEMATICS 



3. Given that the angle, arc sin i is obtuse, construct the 
angle. 

4. Construct the following angles of the first quadrant : arc 
sinf, arc tan (+ 2), arc cos ^-, arc sin .43. Give the approxi- 
mate value of the other two principal functions in each case. 

5. Give five solutions of each of the following : 

arc cos \ — ; arc tan 1 = ; arc sin = ; arc cos 1 = 

6. If arc sin .438 = 26°, what is arc cos .438 ? 

7. What is the value of arc sin — .438 ? Give four answers. 
Give the general formula representing angles which satisfy 
6 = arc sin (- .438). What is arc cos - .438 ? 

8. On the following diagram, regarding the circle as having 
a radius of 100, read the numerical value to two decimal 




places, of the sine, cosine, and tangent of each angle repre- 
sented. Each minor division represents 4 units. 



CHAPTER VIII 

TABLES AND APPLICATIONS 

1. Tables. — The tables of the trigonometric functions are 
computed by processes dependent upon formulas derived in 
the higher mathematics. We have shown the graphical 
method of finding sine, cosine, and tangent, which serves also 
to bring out the fact that the sines of angles from to 45° are 
at the same time cosines of the complementary angles ; simi- 
larly since tan (90° — x) = cot x, it follows that the tangents of 
angles from to 45° are cotangents of the complementary 
angles, from 90° down to 45°. Since tables are given of both 
sine and cosine it is necessary to give values of both functions 
only up to 45°, and similarly with tangent and cotangent. 
Thus sin 26° 10' is found in the table of sines which reads 
down with 26° at the left, and below 10' as given at the top ; 
if the cos 63° 50' is sought we look for 63° at the right of the 
table of sines with the minutes to be read below ; and we find 
that the cosine table is the same as the table of sines, but 
reading up ; this brings us to precisely the same place in 
the tables as sin 26° 10', the complementary angle ; similarly 
sin 63° 50' is sought in the row marked 63 at the left of the 
table and leads to the value which read as a cosine represents 
cos 26° 10'. 

For angles greater than 90° the formulas which we have 
given for related angles are applied. Probably the simplest 
formulas to apply to obtain the functions of obtuse angles are 
the formulas, 

sin (90° -f- x) = cos x, 
and cos (90° + x) = — sin x. 

139 



140 UNIFIED MATHEMATICS 

Thus sin 128° 35' = cos 38° 35' ; 

cos 128° 35' = - sin 38° 35'. 

It is well to note that subtracting 90° from angles greater 
than 100° and less than 200° simply increases the tens' digit 
of the angular measure by one, dropping the hundreds' digit. 

The formulas for 180° + ct, and for 360° — a or — «, are 
used for angles in III or IV. 

Since computation is largely effected by means of logarithms, 
it becomes desirable to have separate tables of the logarithms 
of the trigonometric functions. The sines and cosines of all 
angles are numerically less than 1 and so are tangents of 
angles less than 45° ; hence the logarithms of these numbers 
will have negative characteristics. In the logarithms of the 
trigonometric functions, — 10 is to be annexed to the logarithm 
as given in the table for sines, cosines, and tangents up to 45°. 
Thus log sin 30° is 9.6990 - 10 ; log sin 56° 10' = 9.9194 - 10 ; 
log tan 34° 10' = 9.8317 - 10 ; but log tan 56° 10' = .1737. 

2. Interpolation. — The insertion, by interpolation, of the 
natural and logarithmic functions of angles lying between 
those expressly given in the tables follows precisely the same 
lines as in the corresponding problem in the logarithms of 
numbers. Our tables give these functions for angles increas- 
ing by multiples of 10 minutes ; interpolation enables us to 
compute the functions of angles to minutes ; in using tables 
giving the functions to minutes interpolation enables us to 
compute to tenths of a minute. Note that the assumption is 
always that if angles are read to minutes you compute only to 
minutes ; the tables used should correspond to the precision 
of measurement of the given data. Four-place tables are, 
in general, sufficiently accurate for measurements which are 
made to four places in numbers, and to minutes in angular 
measurement. 

Illustrative problems. 1. Find by interpolation (a) sin 36° 15', 
(b) log sin 36° 16', (c) log cos 36° 18', and (d) log tan 36° 14'. 



TABLES AND APPLICATIONS 141 





Tabular Values - 


— Compare with your tables 




angle 


sin 


log sin 


cos 


log cos 


log tan 




36° 10' 


.5901 


9.7710 


.8073 


9.9070 


9.8639 


53° 50' 


36° 20' 


.5925 


9.7727 


.8056 


9.9061 


9.8666 


53° 40' 




cos 


log cos 


sin 


log sin 


log cot 


angle 



.1 


9 


.3 


.4 


.5 


.6 


2.4 


4.8 


7.2 


9.6 


12 


14.4 



The values to four decimal places of the functions of angles 
between 36° 10' and 36° 20' evidently lie between the values 
which are here given. Thus sin 36° 10' is .5901 and sin 36° 20' 
is .5925, an increase of 24 units of the fourth place ; this 24 
is called the tabular difference. To the ten equal steps of in- 
crease from 36° 10' to 36° 20', by minutes, correspond ten in- 
creases approximately equal to each other, in the sines of these 
angles, making a total increase in ten steps of 24 units of the 
fourth place. The tenths of 24 are respectively, 

.7 .8 .9 

16.8 19.2 21.6 

In adding, as our logarithms are given only to four places, 
we add rejecting tenths, and retaining in the last place the 
nearest unit. Thus for tenths of 24 we use always 2, 5, 7, 10, 
12, 14, 17, 19, and 22. The interpolation does not always give 
the correct result to four places, although in the values of the 
sine the error is always less than 1 unit of the fourth place. 
In the above values of sin 36° IV to 36° 19' as given by the 
addition 2, 5, .7, 10, 12, 14, 17, 19, and 22 units of the fourth 
place to .5901 the first value .5903 should be, to 4 places, 
.5904 ; the error is less here than -^ of 1 % of the value taken. 

1. a. sin 36° 15' = .5901 + ^ of .0024 = .5913. 

Method : Tabular difference is 24 ; to .5701 add .5 of 24 
units of the fourth place. 

b. log sin 36° 16' = 9.7710 - 10 + ^(.0017)= 9.7720 — 10. 
Tabular difference is 17; 10.2 is replaced by 10, and this is 
added in the third and fourth decimal places to 9.7710. 



142 UNIFIED MATHEMATICS 

c, log cos 36° 18' = 9.9070 - 10 - T \ (.0009) = 9.9063 - 10. 
Tabular difference is 9 ; cosine and log cosine are decreasing 

functions ; 7 units of the fourth place must be subtracted. 

d. log tan 36° 14' = 9.8639 - 10 + T 4 o (.0027) = 9.8650 - 10. 
Tabular difference is 27 ; 10.8 is replaced by 11. 

2. Find to minutes, by interpolating, the angle when given 

(a) sin a = .5919, (b) log sin a = 9.7717, and (c) log cot a = 
9.8650 ; find a in each case. 

3. (a) sin a = .5919 ; tabular difference is 24 ; given differ- 
ence .5901 to .5919 is 18 units of the fourth place. Among the 
tenths of 24 find the nearest to 18 ; 16.8 and 19.2, respectively 
.7 and .8 of 24, are equally near and the even number of 
tenths is commonly taken, in such cases, by computers. 

sin a = .5919; a = 36° 18'. 

(b) log sin a = 9.7717; a = 36°10' + T V of 10' (to minutes). 

a = 36° 14'. 
Tabular difference is 17 ; 7 is nearest to .4 of 17. 

(c) log cot a = 9.8650 ; a = 53° 40' + ^ of 10'. 

Tabular difference is 27, a decrease; given decrease is 16; 
among the tenths of 27 the nearest to 16 is 6 ; hence a = 53° 46'. 
Had log cot a been given as 9.8651 — 10 or 9.8649 — 10, the 
angle a would again be given as 53° 46'. 

PROBLEMS 

1. Find the 20 natural trigonometric functions following, 
without interpolation ; time yourself ; limit 6 minutes. 

a. sin 36° 10'. g. tan 70° 30'. 

b. tan 63° 20'. h. sin 28° 50'. 

c. cos 34° 10'. i. tan 16° 20'. 

d. cot 80° 00'. j. cos 8° 40'. 

e. sin 59° 30'. k. sin 157° 10'. 
/. cos 48° 50'. I cos 214° 10'. 



TABLES AND APPLICATIONS 143 

m. cot 141° 00'. q. tan - 64° 20'. 

n. tan 329° 30'.' r. sin 384° 00'. 

o. cos 136° 50'. s. cot 756° 00'. 

p. cos - 28° 10'. t. sin 242° 40'. 

2. Find the logarithms of the above 20 trigonometric func- 
tions, timing yourself. Limit 7 minutes. 

3. Find the following 20 logarithms, interpolating ; time 
yourself. Limit 12 minutes. 

a. log sin 36° 14'. / log cos 48° 57'. k. log sin 152° 15'. 

b. log tan 63° 29'. g. log tan 70° 33'. I log cos 214° 26'. 

c. log cos 34° 14'. h. log sin 28° 51'. m. log cot 141° 05'. 

d. log cot 80° 06'. i, log tan 16° 22'. n. log tan 329° 33'. 

e. log sin 59° 32'. j. log cos 8° 48'. o. log cos 136° 57'. 

p. log cos - 28° 11'. 

q. log tan - 64° 26' 

r. log sin 384° 03'. 

s. log cot 756° 08'. 

t. log sin 242° 44'. 

4. Find the angles less than 90° corresponding to the follow- 
ing 20 logarithms ; no interpolation ; time 6 minutes. 

a. log sin a = 9.6878 - 10 k, log sin a = 9.9499 - 10 

b. log cos a = 9.9954 - 10 I. log cos a = 9.8081 - 10 

c. ' log tan a = 9.4898 - 10 m. log cot a = .8904 

d. log cot a = .5102 n. log tan a = 8.9420 — 10 

e. log cos a = 9.8241 - 10 o. log cos a = 9.9640 - 10 
/. log tan a = 9.7873 - 10 p. log cos a = 9.9757 - 10 
g. log sin a = 9.3179 - 10 q. log tan a = .5720 

h. log tan a = .2155 r. log sin a = 8.9403 - 10 

t. log cos a = 8.9816 — 10 s. log cot a = .0152 

J. log cot a = 9.9341 - 10 t. log sin a = 9.9977 - 10 

5. G-ive in each case another angle which Avould satisfy the 
above relationship, in problem 4 ; e. q. if log sin a = 9.6990 — 10, 
a = 30° or 150°. 



144 



UNIFIED MATHEMATICS 



6. Find the following 20 angles ; interpolate ; time yourself. 

a. log sin a = 9.6881 - 10 

b. log cos a = 9.9955 — 10 

c. log tan a = 9.4861 - 10 

d. log cot a = .5101 

e. log cos a = 9.8228 - 10 
/. log tan a = 9.7879 - 10 
g. log sin a = 9.3200 - 10 
h. log tan a = .2144 

i. log cos a = 8.9912 - 10 

j. log cot a = 9.9358 - 10 

l\ log sin a = 9.9502 - 10 

I log cos a = 9.8092 - 10 

m. log cot a = .8955 

n. log tan a = 8.9492 - 10 

o. log cos a = 9.9645 - 10 

p. log cos a = 9.9753 — 10 

g. log tan a = .5699 

r. log sin a = 8.9404 - 10 

s. log cot a = .0137 

t. log sin a = 9.9978 - 10 

3. Angles near 0° and 90°. 

— For angles near zero, 
from 0° to 2°, the cosines 
vary only from 1.0000 to 
.9994; the cosine function 
to 4 places cannot then be 
used for determination of 
the angle to minutes. Simi- 
larly, of course, the sines 
of angles from 88° to 90° 
vary between the same 
limits. For ordinary pur- 
poses it will suffice to 
avoid the use of the cosine 




Angle 10° in a circle of radius 5 inches 

PM = .868 in. ; arc PA = .873 in.; 

AT= .882 in. 



TABLES AND APPLICATIONS 



145 



in the interval from 0° to 2° or 3° or 4° ; the method of avoid- 
ance is explained below. 

In computing graphically the values of sin 6 and tan 6 even 
with a radius of 10 cm., or of 5 inches, the difference between 
tan and sin becomes too small to read accurately when 6 

.131 r ). For 10° which is .1745 radian, 

sin .1745 r is .1736 and tan is .1763 ; for 5° or .0873 r , 
sin #=.0872 and tan is .0S75 r ; for 1° or .01745 r , sin0=.O1745 
and tan is .01746 or 5 places are necessary to exhibit any 
difference between 0, sin 0, and tan 0. 



is less than — (i.e. 
24 v 



1 




























































































































T 






































































































































ll 






































































ll 
































































ll 
















|0 


1 1 


1 




































mU 











sin < < tan d 

for small acute angles 0, 6 is measured in radians 

Evidently, triangular area OAP < sector OAP < area OAT, 
but the area of the triangle 

OAP = ±OAxMP 

= \ r x r sin $ 
= i-r 2 sin B. 

The area of the sector OAP = | r 2 0, since $ is measured in 
radians, and the area OAT— \ r 2 tan 0. 
Whence, by substituting, 

i r 2 sin < \r 2 Q <\r 2 tan 0. 
sin eos 6 



Whence, as 9 diminishes, — — , lying between 1 and a num- 

sm0 

ber approaching 1, can be made as near to 1 as we please. By 
methods of plane geometry, using 30°, 15°, 7|-°, 3f°, together 



146 



UNIFIED MATHEMATICS 



with 72°, 60°, 12°, 6°, and 3° it can be established that cosf° 
differs from 1 by less than T fo of 1 % ; cos }° = .99991 ; for 
any angle 0, less than |-°, 6 will exceed sin by less than y-L of 
1 % and tan will exceed 6 by less than t ±q of 1 % . Similarly 
the discrepancy between sin and tan 6 for 0, any angle 
less than 3|°, is less than i of 1 %, and 
for any angle up to 8° the difference is 
less than 1 % of either value. 

On the earth's surface ordinary dis- 
tances are regarded as straight lines. 
However for many purposes the deviation 
from a straight line is of importance : 
thus particularly with projectiles of long 
range, the deviation is of vital importance. 
In the figure given if PA represents an 
arc on the earth's surface, PT may be 
regarded as the altitude of a balloon, 
aeroplane, or top of a mountain, and TA 
gives the distance of the horizon. Z TO A 
is equal to the dip of the horizon. AM 
is the drop in the distance twice PA, 
i.e. from T an observer would note, on 
the ocean, the complete disappearance of a 
ship of height AM when the ship is at Q. 
By algebraic process, AT — V2 rh + ft 2 ; 
when h is measured in feet, r in miles, 
and TA in miles, this gives for values of 




Arc PAQ on the 
earth's surface 

2L4, horizon dis- 
tance. 

PT, height of ob- 
server. 



h less than 1.5 miles, AT = a/ 
using 3960 miles as r. 



3 ^, correct to \ of 1 % Check 



PROBLEMS 

1. Given that an observer is at a height of 1000 feet, com- 
pute the distance to the horizon, r = 3960 miles. What is 
the dip of the horizon? Note that the tangent of the dip- 
angle is the horizon distance divided by the radius. 



TABLES AND APPLICATIONS 



147 



2. Find the angle subtended at the center of the earth by 
an arc of length 1 mile, 10 miles, 20 miles. 

3. What is 1° of latitude in miles ? 

4. Degrees of longitude vary in length from degrees on a 
great circle of the earth at the equator to at the poles. 
Find the radius of the small circles on which degrees of longi- 
tude are measured, for 40° north latitude. Where else on the 















































































































































































S' 














x=" 


'<:<: 


s4 


T 




**^ 


\A 








*■*•- 


-^ 












>. 


;>*«■• 








p- .» 




\ 






















— " 


— 


~~ 


f 


f" 


B 










\ 






























> 










\ 










































\ 


























n" 












\ 




















































































Circle of 40° N. latitude 



earth's surface would degrees of longitude be the same ? From 
35° to 45° N. latitude discuss the percentage variation in 
degrees of longitude, as compared with degrees of longitude 
at 40° X. latitude. 

5. How far below the arc of 1 mile on the earth does the 
corresponding chord fall at the lowest point ? Find the same 
distance in inches for arcs of 2 miles, 8 miles, 10 miles, 16 
miles, 20 miles. 

6. What part of the height of a mountain, measured on 
the altitude, is not visible from a point 20 miles distant ? 

7. From what distance can the top of a mountain 10,000 
feet high be seen ? 

8. What distance from shore is a ship whose masts, 
55 feet high, are just disappearing from view ? 



148 UNIFIED MATHEMATICS 

9. Using the figure in the text, find an approximation for 
TA in miles when h is small and measured in feet. 



AT = l 2h '' r \ ( h Y= l 2 - h ' 60 if h V 

\'5280 V52807 \ 80 V5280; 



' 6 « ■ ' h V VL57* 



V 4 ' + V52807 



for values of h less than 5 x 5280, the f- ) can be 

neglected. ^° ^ 

10. Find the " dip " of the horizon and the distance from 
the balloon for h = 100, 500, and 1000 feet. 

11. Find the distance from the point below the balloon on 
the earth's surface to the points on the horizon viewed by 
the observer in the balloon. 

12. According to the approximate formula of Huyghens 
the length of a circular arG, a, is connected with the chord, c, 
of the arc and the chord, h, of half the arc, by the formula 

a = — -jp—. Compute the actual length. 



CHAPTER IX 



APPLICATIONS OF TRIGONOMETRIC FUNCTIONS 

1. Parallel and perpendicular lines. — The slope of the line 
joining (i\, y x ) to (x 2 , y 2 ), 

x 2 — »i P X Q 

evidently represents the tangent of the angle which the line 
joining these two points makes with the .positive ray of the 
cc-axis, i.e. the angle from the #-axis to this line. We have 
taken P 2 ( JB 2j 2/2) to the 
right of Pfa, y{), but 
obviously interchanging 
P 2 (x 2 , yi) and P x {x x , y x ) 
simply changes sign of 
both numerator and de- 
nominator of the fraction 
representing the slope m ; 
Q is in each figure the 
point (x 2 , ?/i) > and P X Q 
and QP 2 have like signs 
if P X P 2 or P 2 P\ makes a 
positive acute angle with 
the positive ray of the 
./•-axis ; and P, Q and QP 2 
have unlike signs in the 
contrary case when P X P 2 X ~ Xi 

or P 2 P X makes a negative acute angle with OX. It is to be 
noted that shifting the y-axis, parallel to itself, either to the 
right or to the left does not affect the value of x 2 — x x , since 

149 

























.h- 








1 








=k-- 


— - 












—^ 





yi)~ 












< 












^ 




-s 





-f» 


Us 


SOT" 


jjt L 






^ 







1— y-^{- — x>{— Xi~ 

r* — 1— — — r~ — — 

ft*-a—\ -j- 1 




^(X 2: 


Ui)zz 




-1 1 1- 




1 






. 


11 : ; : 


^x 2 - 




i 2 


J; 
— 




— ( 


x&kh 


N 









(X-,y, 


— 




— 


y — - 


Y^ 






■ 























































1 'X. 
















- — - 












(Xp 




- 


T 


» -1 


h-y 


1 




' ' 



150 UNIFIED MATHEMATICS 

whatever the position of 0, A X A 2 = P X Q = OA 2 — OA x = x 2 — x x ; 
similarly no change is made in the value of the slope by 
shifting the &-axis parallel to itself, up or down. 

Given y = m x x + k, any straight . line, mj represents the 
tangent of the angle which this line makes with the positive 
ray of the cc-axis. Any parallel line has the same slope ; 
m 2 = m l for two parallel lines. Any perpendicular line has 
the slope angle 

a 2 = 90° + a x ; tan a 2 = tan (90° -\- a l )=— cot a x = — — , 

tan «! 

whence m 2 = — — . Of two parallel lines the slopes are equal, 

and of two perpendicular lines the slope of the one is the 
negative reciprocal of the slope of the other, i.e. 

m 2 = or, by solving, m l = 

?% m 2 

Given y = mx -f- b, any family of parallel lines of slope m. 

y = x + k represents the family of perpendicular lines. 

m 



Illustrative problem. — Given 3x + 4y — 7=0, find the slope, the 
parallel line through the origin, the family of perpendicular lines, and 
the perpendicular line through (— 1, 5). 

4y =-3x + 7. 
y== -%x + %, m = tan<?=- f, 0=-36° 52'. 
y = — f x is the parallel line through the origin. 
Derive this both from y = mx + b and y — y x = m (x — x{) . 

1 4 

The perpendicular line has the slope, m 2 = = -\ 

mi 3 

y = | x + k is the family of perpendicular lines, 
y — 5 = $(x + 1) is tne perpendicular line through (—1, 5). 

EXERCISES 

1. Write the equations of the sides of the triangles used in 
finding the functions of 30°, 45°, and 60°. 



APPLICATIONS OF TRIGONOMETRIC FUNCTIONS 151 



2. Gravity imparts to a falling body a vertical velocity of 
32 t feet per second, with t seconds as time during which the 
body has fallen; on a smooth in- 
clined plane gravity imparts a ve- 
locity of 32 t • sin a where a is the 
angle of inclination of the plane. 
Find the velocity imparted at the 
end of 1 second to a body sliding 
(without friction, assumed) on an 
inclined plane of slope 10°, 20°, 
30°, 40°, 50°, .-.to 90°. 

3. In a freely falling body 



:p 




? 


-\P-- ?- ? * 






j£ai*zl 


i*%** 


j? i 


„* ? V J 


- *.-#' \— \ 


^_^^^y_ MU 1 


C lOv 
















, . , ,. 1 1 . 



Acceleration down a plane, 
g sin ol 



s = 16 t 2 ; while on a plane s = 16 t 2 • sin a ; find s for t = 10, 
a = 30°, 45°, and 60°. 

4. To pull the body up the plane requires a force of 
Wsma + kW' cos a, where k is a constant dependent upon 
the friction. Find the force to pull a weight of 1000 lb. up 
an incline of 30°, Jc = 1 

5. Find the slope of the line joining (—3, 7) to (5, 9) ; 
find the middle point of this line; find the equation of the 
perpendicular bisector of the segment. 

6. Write the equation of the line through (—3, 5) making 
an angle tan~i f Y (m = f 2 -) with the a>axis, and write the 
equation of the perpendicular from (1, 8) to this line. 

7. Find the foot of the perpendicular line found in prob- 
lem 6 and then find the distance between (1, 8) and the origi- 
nal line, using the distance formula. 

8. Find the slope angles in degrees and minutes of the 

following lines : 

(a) 5y- 12 x-T = 0, 

(b) 12y + 5x-3 = 0, 

(c) x - y - 5 = 0, 

(d) Sx-y- 8 = 0. 

9. Find lines through (1, 5) parallel and perpendicular to 
each of the lines in the preceding exercise. 



152 



UNIFIED MATHEMATICS 



— 


X- it 


-X- - _ _ ± 


1o Jr " 


— ^ = ^==r== s =======;#F :: 


__, r _ -Trtr - "" 










,i' T'<r X 






YT :M" ?l 


X± _ ii i 



10. Find the perpendicular bisectors of the sides of the 
triangle formed by the three lines given by the equations, 
5y-12x-7 = 0, 12y+5x-3=0, and x+y-5=0. Find the 
area of this triangle graphically and analytically. 

2. Projections of vectors. — OP has been designated by r, for 
radius vector of the point P. The line OP has magnitude, given 
by r, and direction, given by the angle 0. We may use this 

system of representation to repre- 
sent velocities, forces, and other 
physical quantities. As a velocity 
this vector may be resolved into 
two component velocities, repre- 
sented by OA and OB. OA repre- 
sents the velocity in the x direction, 
x = r cos ', OB represents the y 
velocity, r sin 0, the vertical com- 
ponent of the velocity of a body 
moving with velocity represented by 
OP. The projection of any vector 
upon a directed line is defined as 
the directed distance between the perpendiculars dropt from 
the extremities of the given vector upon the line ; it is given 
by v cos a wherein v represents the vector and a is the 
angle between the positive rays of the two lines. Since 
cos (— a)= cos a, we do 
not need to distinguish 
between the two lines, 
i.e. the angle can be 
taken as obtained by ro- 
tation from the given 
line to the given vector, 
or vice versa. 

It is a fundamental as- 
sumption that any two vector quantities which may be repre- 
sented acting together at the same point may be replaced by 



Components of a vector 

OA represents the x com- 
ponent of OP. 

OB represents the y com- 
ponent of OP. 

OP is the resultant of OA 
and OB. 




Vector parallelogram 



APPLICATIONS OF TRIGONOMETRIC FUNCTIONS 153 



a single vector which is the diagonal of the parallelogram 
formed by the two given vectors. The process is called 
vector addition. This assumes that in space, for example, an 
imparted velocity S. E. of 50 miles per hour increased by a 
velocity N. E. of 30 miles per hour produces the same displace- 
ment whether the two forces which produce the velocities act 
together for one hour, or whether both act in succession each 
for an hour. 

The projection of a broken directed line upon a given 
directed line is the same as the projection of the straight line 
joining the ends of the 
broken line. 

This follows from the 
fact that on a directed 
line 

MiM 2 +m 2 m 3 = MiM b , 

whatever the relative po- 
sitions of Mi, M 2 , and 
M 3 . The directed length 
MiM 2 is the projection of 
PiP 2 , M 2 Mz is the pro- 
jection of P 2 P 3 , MiM z is 
the projection of P X P 3 . 
The physical interpreta- 
tion is simply that the 
total component in the 

x direction (or any other) imparted by two (or more) vectors 
is the algebraic sum of the two (or more) x components of 
these vectors, taken separately. 

When the velocity is given as v, v x and v y are commonly 
used to designate the x and y components of the velocity; 
evidently, also 

v 2 = v 2 + v 2 , 

v x = V cos 0. 

v u = v sin 0. 



II 1 ,..,.. 




: i ! ! : : ! ;_p! 
















S i \ 


\ / V y 




~P\i/* ! V 


r ' \ 


I | \ 






1 ! Il 1 \ 






mi hi! | X 


1 1 ' 1-^3 




1 : : i 1 1 


I 


i 1 1 


; ; 1 1 


1 


1 


1 | 


1 ' ■ ' 1 ! Ill 




1 1 ! ' 1 


i 1 1 ■ 1 ■■! -_!_____ 




i-A/i ~3li~ ~M'~t 


in ii i 3 i 




h . 



Projection of a broken line on a directed 
line 



154 UNIFIED MATHEMATICS 

PROBLEMS 

1. A bullet, muzzle velocity of 3000 feet per second, leaves 
the gun elevated at an angle of 10°. The position, neglecting 
air resistance, is determined at the end of t seconds by the two 
equations : y = 3(m % gin 1Q o _ 16 ^ 

x = 3000 1 cos 10°. 

Find t when y — ; when y = 5 ; explain the two values in 
each case. Find x for both values of t which make y = 0. 

2. The velocity is a vector resolved into components 
v x = v cos a and v y = v sin a. Find v x and v y when a = 10°, 
20°, 30°, 45°, 60°. 

3. A ship sails S. E. for 2 hours at 8 miles per hour and 
E. K E. (22|° off East) for 2 hours at 6 miles per hour. 
Find the x and y of the resultant position. 

4. The propeller imparts to a steamer a velocity of 8 miles 
per hour S. E. (— 45°) and the wind imparts a velocity of 
E. N. E. (+ 22i°) of 6 miles per hour. Find the position at 
the end of 1 hour. 

5. A boy runs east on the deck of a steamer at the rate of 
20 feet per second ; the steamer moves south at the rate of 15 
miles per hour. Find the actual direction in which the boy is 
moving and his total velocity. 

6. Find the velocity in miles per hour of a point on the 
earth's surface due to the rotation of the earth on its axis ; 
find the velocity per second due to the revolution about the 
sun ; compare, and note that the resultant can never be greater 
than the sum nor less than the difference of the two. Take 
values only to 3 significant figures ; 3960 mi. — r ; 93,000,000 
miles as distance from sun. 

7. The United States rifle, model 1917, has a muzzle velocity 
of 2700 feet per second. Find the horizontal velocity of the 
bullet when the angle of elevation is 1°, 10°, 20°, 30°, and 45° 
respectively. 



APPLICATIONS OF TRIGONOMETRIC FUNCTIONS 155 



3. Normal form of a linear equation. — The slope-intercept, 
point-slope, and two-point formulas correspond to the fact 
that a straight line is determined when one point on the line 
[(0, k ) or (x x , y t ) respectively] and the direction of the line are 
given, or when two points are given. A straight line may be 
determined in many other ways ; one method which gives a 
further useful form of the equation of the straight line deter- 
mines the line in terms 
of the length and direc- 
tion of the perpendicular 
from the origin upon the 
line. 

Thus if a perpendicu- 
lar from the origin upon 
a given line is 5 units 
long, and makes an angle 
of 120° with the z-axis 
(positive ray) geometri- 
cally we construct the 
line by constructing the 

ray of 120° and upon it taking a length of 5 units. At the 
extremity of this line of 5 units length a perpendicular is 
drawn which is the required line. The point N is readily 

found to be (5 cos 120°, 5 sin 120°) and the slope is 
therefore the equation of the line to be found is 



.. . I 






A * 


,tt 




s* '- 


% ^f - 




. r ^M v 






-^>rV^ V-> rv 




_'v>'M<. ( V \* ->& 


V A> ,Jnk '\ : y -£a 


IsVEPv^ V \ 


..(Oh^C>X V \ 


Vl**ei' \ ■- 


-£--£- S 










__:___ 



A line determined by the normal to it 
from the origin 

Normal length, 5 ; a — 120°. 



-1 



tan 120 c 



5 sin 120° = 



tan 120 c 



(x - 5 cos 120°). 



sin 120° 
tan 120° = - : — -; substituting, clearing of fractions, trans- 



posing 



cos 120 c 

x cos 120° + y sin 120° - 5 (sin 2 120°+ cos 2 120°)= 0, 
x cos 120° + y sin 120° — 5=0, since sin 2 a + cos 2 a = 1, 
1 ,V3 - n 

-2 x + -Y y - 5 = ' 



156 



JUNIFIED MATHEMATICS 



In general, given the normal ON to the line from the origin, 
of length p, and making angle a with OX, the extremity N is 

(p cos a, p sin a) ; the slope is 

, and the equation becomes 



m 



% 



5 



cos a 



sin a 
x cos a 4- y sin a — p = 0, 



ij+ttj: p is taken as a positive quantity 
just as r has been taken. Evi- 
dently if p = 0, 



a? cos ft + y sin a = 



represents a parallel line through 
the origin. Evidently also for 
parallel lines on opposite sides of 

the origin the angles a and a' differ by 180° ; i.e. a' = 180° + a, 

whence 



x cos CL + y sin ft — p = 

Normal form. 



Sill a' — — sill ft. 



COS ft' 



COS ft. 



::::::±:::::::::::::::::::::::::^^::|SS 

;; P1»|1 ll 

= ========= = H^==|E-g::E=E=|=:E=|==E=E: 

==i^=== ====::==Ig==:===::===:=:tEEEEEE 

:"""E""E#xE££"""±E±±fei/rES 
Tffff ^ Tffffi^^aTffl 1 1 1 1 1 1 1 1 1 1 i M I T T 



The projection on ON 1 j the projection on ON 

of OM + MP j e <J uals j ofOP 

4. Normal form derived by projection. — We have shown that 
the projection of any broken line upon any given line is the 



APPLICATIONS OF TRIGONOMETRIC FUNCTIONS 157 

same as the projection upon the given line of the vector join- 
ing the ends of the broken line. Let P(x, y) be any point 
on the line whose equation is sought; drop PM the perpen- 
dicular from P to the a>axis ; the projection of the broken 
line OM+ MP on the normal OAT is equal to the projection of 
OP on ON. Now OM = x makes the angle a, by hypothesis, 
with ON, and MP makes the angle a — 90° ; hence the projec- 
tion of OM on ON is x cos a (OA, negative in the figure 
since a is obtuse) and of MP on ON (AN in the figure) is 
y cos (a — 90°) ; the projection of OP on ON is ON itself, or p; 
further y cos (a — 90°) = y cos (90° — a) = y sin a. Then, since 
projecting on the line ON, 

projection of OM + projection of MP— projection of OP, 
we have 

x cos a + y sin a = p, whence x cos a + y sin a — p = 0. 

5. To put the equation of a straight line in normal form. — 

Let the given equation be 3 x — 4 y + 7 = 0, and let 

x cos a 4- y sin a — p = be the same equation in normal form. 

If these two equations represent the same line, these lines 
must have the same slope and the same y (or x) intercept. 



- cos a 


= 


3 


i 


: P 


sin a 




4' 


4~ 


sin a 


cos a 


= 


^4 


3 . 

- sin a. 


cos 2 a 


= 


9 
1 6 


sin 2 


a. 



But cos 2 a = 1 — sin 2 a, 

whence 1 — sin 2 a = T 9 6 sin 2 a ; f-f sin 2 a = 1 ; sin a = ± |. 
j9 = + ^ sin a, whence since p is to be positive, sin a must be 



4. 40 = X • POS tt = — . ^ J 



taken as positive. . Hence sin a = + f, p = -J ; cos a 
and thus the normal form is — ^x + ±y — -J = 0. This equa- 
tion is obtained by dividing each member of the original 
equation by — 5. 



158 



UNIFIED MATHEMATICS 



In general to put Ax + By + C — in the normal form, 
x cos a + y sin a — p = 0, one must multiply through by some 
quantity k, so that k A = cos a, kB = sin a, and kC = —p; 
kC=—p shows that k must be chosen opposite in sign to O; 
squaring both members of the first two equations and adding 

gives k 2 (A 2 + B 2 ) = 1, whence k = ± — , of which the 

sign is taken as opposite to C. vA+B 

Eule. — To put an equation Ax + By + C = in normal form 
divide through by = ± V 'A 2 4- B 2 , with the sign taken opposite to 
that of the constant term. 



~A 



n 



-A 



m 



m 



tt 



t 



M 



m 



IMS 



6. To find the perpendicular distance from a point to a line. — 

In solving this problem one considers the various forms of the 

straight line 
which may be 
employed. Evi- 
dently the normal 
form is most 
hopeful for use, 
since it involves 
the-perpendicular 
distance of the 
given line from 
the origin. 
Through the 
point P x (x x , y x ) 
draw a line paral- 
lel to the given 
line ; evidently 
the difference be- 
tween . the nor- 
mals to the two 
lines gives the 
distance. Three possibilities must be considered : 1. P 1} on 
the opposite side of the given line from the origin; 2. P u 



Distance of a point from a line 



APPLICATIONS OF TRIGONOMETRIC FUNCTIONS 159 

on the same side of line as 0, the origin, but such that the 
normal angle is the same, i.e. so that the parallel line through 
P 1 (x 1 , y x ) falls on the same side of as the given line, P\ on 
the figure ; 3. P u on the same side as the origin, but the 
normal angle increased (or diminished) by 180°, designated by 
P'\ on the figure. 

Let x cos a + y sin a — p = be the equation of the line. 

1. x cos a + y sin a — fa cos a -+- Vi sin a) = is the parallel 
line through Pi fa, y x ), since this equation is evidently in 
normal form and the line passes through fa, y x ). 

ON 2 = Xi cos a + 2ft. sin a - 
d — ON 2 — ON = x x cos a -f 2/x sin a— p. 

The perpendicular distance is obtained then by writing the 
equation in normal form and substituting for fa, i/i) the 
coordinates of the given point. Evidently if Pi(x, y) is on 
the line, this gives also the correct distance, which is then zero. 

'2. x cos a+y sin a — fa cos a+y^ sin a) is the equation of the 
parallel line ; again, ON' 2 = a?i cos a -f- 2/i sin a. 

d = OxV - 0^2 : whence - d = ON' 2 - ON 

= Xi cos d-\-yi sina — p. 

The same rule holds, but the distance in this case is negative. 
Evidently the rule holds if ON' 2 is 0. 

3. a' = 180 + a ; cos a' = — cos a, sin a'= — sin a. 

To write the equation of the parallel line in normal form, the 
coefficients of x and y must both be the negatives of the coeffi- 
cients of x and y in the given equation. 

x ( — cos a) -\-y(— sin a) — ( — fl?i cos « — y x sin «) = is the 
equation of the parallel line in normal form. 

ON" 2 = — fa cos a + y x sin a). 
d = OiV" 2 + OiV= — »! cos « — y x sin a +79, 
or — d = #! cos a -+- y x sin a — ;;. 



160 UNIFIED MATHEMATICS 

Rule. — To obtain the distance from a point to a line write 
the equation in normal form, substituting therein for x and y the 
coordinates of the given point. The resulting number gives the 
distance as positive if the point and the origin lie upon opposite 
sides of the given line, as negative if Pi and are upon the same 
side of the given line. 

x cos a + y sin a — p represents the perpendicular distance 
then from P (x, y) to the line x cos a + y sin a — p = 0. For 
all points on one side of this line the expression is positive, 
and on the other side, crossing the line to the origin side, the 
expression is negative. 

A line which passes through the origin, p = 0, will be said 
to have its equation in normal form when sin a is taken as 
positive, i.e. when the coefficient of y is made positive. Points 
on this line make x cos a + y sin a = ; points above the line 
make the expression x cos a + y sin a positive, and points be- 
low the line make it negative. 

Thus Sx— 4 1/ = is written ^— x + - y = 0, or 

5 5 

_3a; + 4 ?/ _ 
" 5~ ~ U ' 

to be in normal form. The perpendicular distance from any 
point to such a line will be positive for points above the 
line, and negative for points below the line. 

7. Bisector of the angle between two lines. — Geometrically 
the bisector of an angle is the locus of the points equidistant 
from the two sides of the angle ; analytically we express the 
condition that two distances should be equal to each other. 
Let the equations be given in normal form, as 
x cos «! + y sin a x — p 1 = and x cos a 2 + y sin a 2 — p 2 = 0. 

Let P(x, y) represent any point on either bisector of the 
given angle ; analytically 

x cos «! + y sin a { — p L = ± (x cos a, -f y sin « 2 — ]) 2 ). 



APPLICATIONS OF TRIGONOMETRIC FUNCTIONS 161 

















}" 
































,s>. 






/'-* 


















"4p- 




• 




















,d 


A/>-s£ 








B 














>y 


7<5 x i^5s 


H 














/ 










% 


■ 






//■ 
















\ 




\ 


V- 


// 


"> 












A" 






\ 


\ 




fyS 


P C 




















n) 
























■N' 


/ 


fe 




\ 




















*' 








\ 


















X 


M 








\ 


s 
















1 

























Bisectors of the angles between two 
lines given normal form 



Bisector ^1, in the opening which, includes the origin, is ob- 
tained by taking the + sign since both perpendiculars are of 
the same sign for points on A. Bisector B is obtained by 
taking the negative sign since 
any point on B is on the 
same side as the origin with 
respect to one of the lines, 
and on the opposite side with 
respect to the other ; hence 
if PM 1 comes out negative, 
PM 2 will be positive (by the 
formula) and the equality 
will be obtained by putting 
PM X = - PM 2 . 

Just as the two axes divide 
the plane into 4 quadrants in 
which the distances to these 
axes are + +, — +, — — , 

and + — respectively, so any two lines in the plane divide 
the plane into 4 sections in which the perpendicular distances, 
as given by our formula, to these lines are + +,+ —>——> 
and — + respectively. The + + and — — sections are 

separated by the H and h sections respectively, as it is 

evident that you pass from + + to + — by crossing the 
second line. 

The bisector of the + -f and — — opening is given by 
equating the left-hand members of the equations of the two 
lines in normal form ; the bisector of the + — , — + opening 
is obtained by equating the one to the negative of the other 
left-hand member. 

If one of the lines passes through the origin, or if both 
do, then the above-mentioned convention is necessary to es- 
tablish the part of the plane in which the left-hand member of 
the equation of the line is positive. It is customary to make 
sin « positive, which makes the portion of the plane above the 
line the positive side, i.e. the coordinates of any point above 



162 UNIFIED MATHEMATICS 

the line when substituted in the given equation give a positive 
value, and of any point below the line give a negative value. 

PROBLEMS 

1. If a line makes an angle of 30° with the a>axis what 
angle does the normal to the line make with the se-axis ? 

2. What is the slope of the line, y = 2 x + 5 ? What is the 
slope angle ? What is the slope of the normal to this line ? 
What is the angle which this normal makes with the cc-axis ? 
Find from the tangent of the angle made by the normal with 
the o>axis the sine and cosine of the same angle. Write the 
equation in normal form and interpret the constants. 

3. Given a and p, as below, slope angle of the normal and 
length of the normal from the origin to the line, find the 
equations of the lines, and draw the lines : 

p — 5. 
p = 5. 
p = 5. 
p = 5. 
p = 5. 
p = 10. , 
p = 10. 
15', p = 8. 

4. If a remains equal to 40° and p varies, what series of 
lines will be obtained ? if p remains equal to 5, and a varies, 
what series of lines will be obtained ? 

5. Write the following equations in normal form : 

a. 3x — 4?/ — 5 = 0. 

b. 5x + 12^ + 7 = 0. 

c. 5x-\-12y-7=0. 

d. 3x-5y-± = 0. 

e. y = 2 x — 14. 

£ 3y-7x+52 = 0. 

9- 1 + 1=1. 



a. 


a = 30°, 


b. 


a = - 30°, 


c. 


a= 150°, 


d. 


a= 210°, 


e. 


a= 137°, 


f. 


a= 137°, 


9- 


a - - 63°, 


h. 


a= 223° 



APPLICATIONS OF TRIGONOMETRIC FUNCTIONS 163 

6. Find the distances of the points (1, 5), (2, 3), (0, 5), 
(0, — 5), (—2, —3), and (—3, 7) from each of the lines in 
the preceding problem. 

7. What is the distance of the point (x, y) from the line 
Sx—Ay — 5 = 0? Under what circumstances does the formula 
give a negative value for this distance ? What is the distance 
of any point (x, y) from 5x + 12y-\-8 = 0? What does 
equating these two expressions, i.e. the left-hand members of 
each normal form, give ? Interpret on the diagram. What is 
obtained by setting one of these expressions equal to the 
negative of the other? 

8. Find the bisectors of the angles between the following 
pairs of lines : 

a. 3x + ±y-5=0sin&12x-5y-10 = 0. 

b. y—2x—5=0 and 2x + y + 7 = 0. 

c. y — 2 x — 5 = and 3 y + x — 8 = 0. 

d. #-2a,- = 0and3y + a;-8 = 0. 

e. y — 2 x = and 3 y + x = 0. 

9. Find the distance of the points (1, — 3), (3, 0), (3, — 7), 
and (0, — 8) from each of the lines in the preceding problem. 

10. Find the distance between the following pairs of paral- 
lel lines : 

a. y = 2 x — 7, 

y = 2x + 3. . 

b. 4 y — 3 x = 5, 
4y-3x-16 = 0. 

c. 4 y — 3 x = 0, 

4 y - 3 x — 16 = 0. 

d. x + 2y — 7 = 0, 

2x + ±y + 17 = 0. 

e. 7.2z + 8.3 y -15 = 0, 
7.2x + 8.3y- 8 = 0. 



164 UNIFIED MATHEMATICS 

11. In problem 8 show that each bisector obtained is one 
of the pencil of lines through the point of intersection of the 
given two lines. 

12. Find the area of the triangle having as vertices the 
following points : 

a. (3, 4), (0, 0), and (0, 8). 

b. (3, 4), (0, 0), and (10, 2). 

c. (1, 1), (4, 5), and (7, - 3). 

13. Find the area of the triangle formed by the three lines : 

3^ + 4?/- 5 = 0, 

12 x - 5 y - 10 = 0, 

and 4 x — 3 y — 7 = 0. 

14. What is the distance of any point (x, y) from the point 
(0, 0) ? What is the distance of any point (x, y) from the 
line x — 5 = ? Equate these two expressions for distance 
and simplify. The resulting equation has for its graph all 
points which are equally distant from the point (0, 0) and the 
line x — 5 = 0. 

15. Find the locus of all points which are equidistant from 
the point (0, 0) and the line y — 8 = 0. Let (x, y) represent 
any point satisfying the given condition. 

16. F'ind the locus of all points at a distance 10 from the 
point (0, 0) ; from (1, — 3). Find the locus of all points at a 
distance 10 from the line 3x— 4?/— 7 = 0; at a distance — 10 ; 
explain graphically. 

17. Find the locus of all points equally distant from 
3 x — 4 2/ — 5=0 and from (1, — 5). 

.18. In problem 12 find the equations of the three bisectors 
of the angles of the triangle formed ; find the perpendiculars 
from the vertices to the opposite sides ; find the perpendicu- 
lar bisectors of the sides ; show that in each instance you 
have three lines which have a point in common. 



APPLICATIONS OF TRIGONOMETRIC FUNCTIONS 165 

19. What points, when the coordinates are substituted fol- 
ic and y, make the expression 4 y — 3 x — 5 positive ? What 
points make this expression zero? What points make this 
expression negative ? Locate three points of each type, plot 
and discuss. 

20. Substitute in the expression x 2 + y 2 — 25 for x and y 
the coordinates of the points (0, 3), (± 3, 2), (± 1, 4). Plot 
these points. Substitute (0, ± 5), (± 3, ± 4), (± 4, ± 3), and 
(± 5, 0). Plot. Substitute also (0, 8), (±7, 0), (5, 3), and 
(± 4, 6). Xote that the graph of x 2 + y 2 — 25 = separates 
the plane into two parts ; in the one part inside this curve are 
all points whose coordinates substituted for x and y, respec- 
tively, make the expression x 2 +' y 2 — 25 negative, and in the 
part outside lie all points which make this expression positive. 

21. Prove that the perpendicular bisectors of the sides of 
any triangle meet in a point ; the. vertices may be assumed as 
Oi> 2/i), (^2, y 2 ) and (x 3 , y s ) or as (0, 0), (x 1} 0), and (x 2 , ;y 2 ). 

22. Prove that the bisectors of the angles of any triangle 
meet in a point. 

23. Given the three vertices of a parallelogram, how do you 
find the fourth vertex? Apply to (1, 5), (6, - 1), and (3, 2). 

24. How do you find a line parallel to a given line at a given 
distance from it ? 



CHAPTER X 

ARITHMETICAL SERIES AND ARITHMETICAL 
INTERPOLATION 

1. Definition of an arithmetical series. — In the table of natu= 
ral sines the valnes of the sines of 21° to 22° are given as 
follows, 

sin 21° sin 21° 10' sin 21° 20' sin 21° 30' sin 21° 40' sin 21° 50' sin 22° 
.3584 .3611 .3638 .3665 .3692 .3719 .3746 

It is to be noted that each value differs from the preceding by 
.0027, and each angle differs from the preceding by 10'. 
Either of these sequences of numbers with a constant differ- 
ence between each number and the preceding is termed an 
arithmetical series ; the continuation of the lower series by the 
successive addition of .0027 gives indefinitely further values 
of the arithmetical series, but gives only 5 following sines. 
Of the series of numbers giving to four de'cimal places the 
values of the sines of angles which increase by intervals of 10' 
it happens, for reasons which will be further discussed below, 
that twelve values beginning with the sine of 21° coincide with 
the first twelve terms of an arithmetical series ; the sine 
series must not be confused with the arithmetical series, as it 
is only arithmetical in limited intervals and then only when 
approximate values are used. Thus if five place values of the 
sines of the angles above were given the series would no 
longer be arithmetical. 

The type form of arithmetical series is 



a a ■+ d 


a + 2d- 


a + Sd 


.. a + 9d 


... a+(n-2)d 


a+(n - l)d 


1st term 2d 


3d 


4th 


10th 


.. (n-l)thterm 


nth term 



each term is d greater than the preceding term of the 
series. 

166 



ARITHMETICAL SERIES AND INTERPOLATION 167 

l n = a + (n — l)d ; by l n we designate the nth term of such 
a series. It is evident from the definition that the tenth 
term in such a series is a -f 9 d, since the common difference 
d appeared first in the second term and one further d was 
added in each subsequent term. 

2. Last or nth term, and sum. — Strictly we should prove by 
a process called mathematical induction, that the formula, 

l n = a + (n — l)d, 

always represents the nth term. Evidently for n = 1 this 
does represent our first term ; for n = 2 the expression does 
represent our second term ; for n = 3 the expression a + 2 d 
does represent our third term ; let us suppose that for n this 
does represent our ?ith term, then our (?i -f- l)th term, which is 
d greater, must be l n -f d = a + (n — l)d -f d = a + nd ; now 
the formula gives Z, )+1 = a +.(?i + 1 — 1)^ = « + ^^ ; hence 
if this formula is correct for the nth term, the formula is cor- 
rect for the next, the (n -\- l)th term. However, we know that 
the formula is correct for the third term, hence it is, by our 
theorem just stated, true for the next, the fourth term ; since 
it is true for the fourth it is, by the theorem, true for the fifth ; 
so for every subsequent term. 

Frequently the sum to n terms, s n , of such a series is desired. 
To obtain a simple expression for s n , Ave proceed as follows : 

s n = a + (a + d) 4- (a + 2 d) + ••• a + 9 d + ••• a + (n — l)d 
or l n ; reversing the series gives, 

s n = l n + Q n -d) + ft - 2 d) + -. l n - 9 c? + -. Z„ - (n - l)d ; 
adding, 

2 s„ = (« + g + (a 4- y + (a + U + •••(« -f U + • • • (a + ZJ ; 

Fundamental formulas : 

l.=o + ("-lK 



168 UNIFIED MATHEMATICS 

3. Practical importance. — Arithmetical series are of great 
importance because of their occurrence in practical problems, 
and because they are fundamental in the applications of 
mathematics to statistical problems. In physical problems 
involving time, the time is commonly measured at the end of 
equal intervals, giving an arithmetical series for the time ; in 
the tables of logarithms our numbers increase arithmetically, 
and so in the tables of trigonometric functions the angles 
increase arithmetically. Kefmeinent of measurement is com- 
monly made by subdividing the unit of measurement into 
smaller equal intervals, giving new arithmetical series. 

PROBLEMS 

1. Find the tenth and the twentieth terms of the series, 
1, 3, 5, 7, ... ; find the (n -f- l)th term. 

2. Find the sum to 10 and to 20 terms of the series 
1,3,5,.-. 

3. Show by mathematical induction that the sum of the first 
n odd numbers is n 2 , by showing that if the sum is n- then the 
sum of the first (n + 1) odd numbers is (n -f l) 2 . 

4. Solve I = a + (n — l)d, for d ; solve for a j solve for n. 

5. Given I = 235, d = 7, n = 40, find a. 

6. Given I = 235, n = 7, a = 5, find d. 

7. Given d = 7, a = 5, n = 40, find I. 

8. One hundred men increase uniformly in height from 5.01 
feet to 6 feet by .01 of a foot, find the total height ; if their 
weights increase uniformly by half-pounds from 110 pounds, 
find the total weight of the group, and the average weight. 

9. On an inclined plane, angle of 30°, a ball rolls approxi- 
mately 8 feet in 1 second, 24 feet in the second second, 40 feet 
in the next, and in every second 16 feet more than in the pre- 
ceding second. Find the distance a ball travels in 5 seconds ; 
in 10 seconds. This formulation neglects the energy-loss due 
to rolling. 



ARITHMETICAL SERIES AND INTERPOLATION 169 

10. On a hill inclined at 30° a bob-sled moves approximately 
according to the law of the rolling ball in the preceding 
problem. Find the length of time to cover 1000 feet. Find 
the distance covered in the last second of the slide. Find 
the average velocity during the slide, and the average velocity 
Reduce velocity to miles per hour. 



during the last second 



Note. — Average velocity is simply the space covered divided by the 
time required. 



4. Graphical representation. — The arithmetical series is 
represented graphically by the straight line, and conversely 
any straight line represents an arith- 
metical series. For this reason the 
interpolation processes explained 
above under logarithms, and under 
trigonometric functions are some- 
times termed " straight-line interpola- 
tions " ; the process is correct in those 
small intervals in which the curve 
representing the function is approxi- 
mately a straight line. 

For the integral values of x, from 
0, 1, 2, 3, ---up to n — lj the ordi- 
nates of the line 

y = dx + a 
represent graphically the terms of 
the type arithmetical series, a, a + d, 
a + 2 d, • • ; for values of x from 0, 
n-1 



_1 2 3 4 

10) 1 0' 10? 10' 



up to 



10 



, the or- 

















Y 




\w 










A'l 










/-l ! 










M I 

/ -i ! 








Cs 


k\\ 








.^/r i ! i i 






\fl[\ d \ j 






\f\d\ ill 






JU t i 1 i 1 

n i ' i ' 






m l 






hd 


it: i 




A l 


— 1 


Mil 

llii 






ay 


imm! 




1 




ihii! 







12345678 X 



dinates represent the terms of an 
arithmetical series with first term a 



The ordinates of y = dx + a 
&t x = 0, 1, 2, 3, ••• repre- 
sent terms of an arithmet- 
ical series 



and the common difference — . To 

10 

any series of equal increases or increments given to x there 

correspond a series of equal increments given to the ordinates ; 



170 



UNIFIED MATHEMATICS 



this depends upon the theorem of plane geometry that if a 
series of parallel lines cut off equal parts on one transversal 

they do on every transversal, and 
this theorem is equally true for 
any straight line in the plane. 



5. Interpolations in sines and 
other functions. — The value of the 
sines of the angles are given by 
the corresponding ordinates in a 
circle of radius unity, or the or- 
dinates divided by 100 in a circle 
of radius 100, or the ordinates di- 
vided by 1000 in a circle of radius 
1000. 

On our diagram, with radius 
100 the straight line joining the 
end of the ordinate corresponding 
to 20° to the end of the ordinate 
at 30° does not differ materially 
from the circular arc connecting 
these points. Were we to plot 
these angles in a circle of radius 
1000 the points of intersection 
would appear as in the second part 
of the diagram, lettered AB, and 
constituting a tenfold linear en- 
largement of AB. 

If the sines of the angles were 
given by intervals of ten degrees, interpolation by tenths would 
give the sines by degrees ; the circle with radius 100 mm. (or 
100 twentieths of an inch) permits the sine and cosine to be 
read to two places accurately, and this rather low degree of 
refinement corresponds to a table of sines given by intervals 
of ten degrees ; interpolation would give substantially correct 
values to two decimal x>laces, e.g. for the sines of 21°, 22°, 



12 3 4 n-m 

Graphical representation of the 
sum of an arithmetical series 

2 



ARITHMETICAL SERIES AND INTERPOLATION 171 



100 




Arc of 20 J to 30° in circles with radii 50, 100, and 1000 fortieths of an inch 

The marks on the long chord indicate the points given by interpolating 
between sin 20° and sin 30°, and between cos 20° and cos 30°. Even on 
the arc with 25 inch radius nine interpolated points on the chord and cor- 
responding points on the arc, between 21° and 22°, coincide. 



172 UNIFIED MATHEMATICS 

..-29° by interpolating between sin 20° = .34 and sin 30° = .50. 
The arc of 10° on this circle differs slightly but appreciably to 
the eye from the chord of 10°, but the interpolated points on 
the chord are not easily distinguished from the ten points 
on the curve. 

Angles given by degrees permit interpolation by intervals of 
6' or by intervals of 10', with substantially correct values to 
the third place if the values are given only to three places ; 
values given to four places give by interpolation values 
substantially correct to the fourth place. On our circle with 
radius 1000 the sine and cosine can be read to three decimal 
places ; interpolation between the values of sin 20° and sin 30° 
give points markedly different from the true points on the 
curve. These points are indicated by checks on the chord of 
10°. Interpolating five points (for 10', 20', 30', 40', 50') on the 
chord from the 20° point to the 21° point gives points not 
readily to be distinguished from the correct points on the arc. 
On this diagram it is not possible to distinguish the sub- 
divisions for minutes on the arcs from the corresponding 
points on the chords of central angles of 10'. 

With the proper changes, noting particularly that as in- 
creases cosine decreases, the argument given holds for inter- 
polated values for cos 0. 

Interpolation of the tangent values is similar, except in the 
neighborhood of 90° where the tangent changes very rapidly ; 
in a separate table are given by minutes, the tangents of angles 
from 88° to 90°. 

The graph of the function y = log 10 x, or 10 y = x, is a con- 
tinuous curve which for small arcs approximates a straight 
line. Similarly the graphs of the functions y = sin x, and of 
2/= sin x, log cos x, log tan x and log cot x approximate straight 
lines within small intervals, and so are subject to our ordinary 
process of interpolation. 

6. Arithmetical means. — If two numbers a and b are given, 
the arithmetical mean between the two is the number x which 



ARITHMETICAL SERIES AND INTERPOLATION 173 

makes a, x, b three consecutive terms of an arithmetical series ; 
to insert n arithmetical means it is necessary that a, the n 
means, and b form n + 2 consecutive terms of an arithmetical 
series. Ordinary interpolation is the insertion between two 
tabular values of some particular one of 9 arithmetical means. 
If a, x, b form an arithmetical series, 

b — x = x — a, 

whence x = — — — 

2 

If a, a + d, a -f 2 d, a -f 3 d, ••• a +(n — l)d, a -f nd, b form 

an arithmetical series, b = a -\-(n + l)d ; whence d = . 

n + 1 

The sum of n terms of the series a, a + d, a + 2 d, • • • 
a + (?i — l)d, is 

the average value of these ?i numbers is 

n (a + ?„) _ a + ?» 

2 • n ~ 2 ' 

termed the arithmetical mean of the n numbers ; the sum of 
an arithmetical series is seen to be the " average value " mul- 
tiplied by the number of terms. Similarly of any collection 
whatever of n quantities, the arithmetical mean is regarded as 
the total sum divided by the number of quantities. In statisti- 
cal work the latter mean, total sum divided by the number of 
given quantities, is called the " weighted mean." 

PROBLEMS 

1. Between .3584 and .3746 insert 5 arithmetical means; 
if .3584 = sin 21° and .3746 = sin 22°, what do these means 
represent? Between .3746 and .3584 insert 5 arithmetical 
means ; interpret as cosines. 

2. Given sin 21° = .3584 and sin 21° 10' = .3611, find 9 
intermediate values ; interpret. 



174 UNIFIED MATHEMATICS 

3. Given sin 0° = 0, sin 30° = .5000, what value would 
arithmetical interpolation give for sin 21°? What is the 
error ? 

4. Given sin 20° = .3420 and sin 30° = .5000 ; find to 4 
places sin 21°. How many terms in the arithmetical series 
which is implied? 

5. What is the sum of the first ten integers ? 

6. If cards are marked 1 to 190, what is the total sum ? 
What is the average value of the total group of numbers ? 

7. How many years of life have been lived by a group of 
30 individuals, aged 21, 22, 23, . . . 50 years? 

8. Falling from rest a body falls approximately 16 feet in 
the first second, and 48 in the second, and in each succeeding 
second 32 feet more than in the one which precedes. What 
distance will the body fall in 10 seconds ? How long will it 
take such a body to fall 1000 feet ? 

9. If it takes a lead ball 8 seconds to fall to the earth from 
a balloon, what is the height of the balloon ? 

10. How long will it take a ball to reach the earth if 
dropped from the top of the Washington monument, 550 feet 
high? 

11. Draw figures to show that between x — o and x = 4, 
the graph of xy = 1 approximates a straight line. 

12. Write the equation of a straight line representing for 
integral values of x, from to 10, the arithmetical series with 
a = 10, d = — 3. Represent the series also by the series of 
rectangles, each of width 1, In summing the arithmetical 
series we reversed the series and added ; show the geometrical 
equivalent on the figure, page 170, with the rectangles. 

13. Given that the first term of an arithmetical progression 
is 8 and the last term 100, what equation must n and d satisfy ? 
If d = 2, what is n ? If d = 3, what is n ? Interpret. 



ARITHMETICAL SERIES AND INTERPOLATION 175 

14. Show that in an arithmetical series of n terms (a, 
a 4- d, •••) the average value is -J the sum of the first and last 
term. Note that the average value of the terms is the total 
sum divided by the number of terms. This average value is 
termed the arithmetical mean of the n terms. 

15. Find the average value of the following 10 heights, and 
the arithmetical mean : 

40 6 feet 126 5 feet 9| inches 

64 5 feet 11| inches 138 5 feet 9 inches 

86 5 feet 11 inches 120 5 feet 8£ inches 

92 5 feet 10| inches 112 5 feet 8 inches 

142 5 feet 10 inches 80 5 feet 7| inches 

16. If there are 1000 men measured and they are grouped 
in height as above, find the average height of these men. 
Note that the easiest way to find this average is to take the 
variation above and below some one of the " middle " values, 
e.g. with reference to 5 feet 9 inches as origin, 6 feet is re- 
garded as + 3 inches and this group has a total of 120 inches 
excess above 5 feet 9 inches per individual ; 5 feet 1\ inches is 
regarded — \\ inches and the total group of 80 has a total 
deficiency of 120 inches, or — 120 inches ; the two neutralize 
each other. Whatever total remains is divided by 1000 and 
added, algebraically, to the 5 feet 9 inches. 

17. Draw the graph of I = a + (w — l)d ; assuming a and d 
as constants. 

18. Historical problem. — In the Egyptian manual men- 
tioned above, occurs the following problem : If 100 loaves of 
bread are divided according to the terms of an arithmetical 
series among 5 people so that A- of what the first three receive 
equals what the last two receive, find the number received by 
each person. Solve the problem. The Egyptian reckoner 
assumes that the last person receives 1 loaf, and without any 
explanation, that the second receives 6^ loaves, and so on in 
arithmetical progression ; the sum he finds to be 60, and to 
arrive at the correct values all numbers are increased in the 
ratio of 100 to 60. Compare with your solution. 



CHAPTER XI 

GEOMETRICAL SERIES AND APPLICATIONS TO 
ANNUITIES 

1 . Geometrical series. — A series of terms in which each term is 
obtained from the preceding by multiplying by a fixed number 
is called a geometrical series ; by definition the ratio of each 
term to the preceding is a constant. Designating this ratio 
by r, and the first term by a, the type series becomes : 

a, ar, ar 2 , ar 3 , ar n ~ l . 

l n = ar n ~ l . 

The sum of such a series to n terms is obtained as follows : 
Let s n = a + ar -+- ar 2 -\- ar 3 -\- ... -h ar n ~ l . 
rs n = ar + ar 2 -f- a?* 3 -f- ... -f- cw n-1 + ar n . 
s n — rs n = a — ar n . 
s n (l — r) = a — ar n . 

_ a — ar n _ a ar n _ ar n — a 

1 — r 1 — r 1 — r r — 1 

2. Sum to infinity, r< 1. — When r is numerically less than 
1, the terms of a geometrical series become smaller and 
smaller without limit. The sum of n terms differs from the 

fixed quantity, , by the quantity ■-, which decreases 

1 — r 1 — r 

as n increases ; this difference between - — — and 



1 — r 1 — r 1 — r 

can be made smaller than any assigned quantity however 

small by taking n sufficiently large ; the value - — — is termed 

1 — r 

176 



GEOMETRICAL SERIES AND ANNUITIES 177 

the " sum to infinity " of the series or, more strictly, the 
" limit " of the sum of the series as the number of terms in- 
creases indefinitely. 

A simple and familiar illustration of a geometrical " sum to 
infinity " is found in the recurring decimals of elementary arith- 
metic. Thus .33333 ..., or .3, is the series -^ H-yf-o + T -oVo + - 
with r = y 1 ^-, a = T 3 ¥ ; .i7 represents the series -£fo + yoVVo 

17 1 7 
_i_ 17 |_ with r — -J— a — - 1 - 7 - s = 1 00 = — 

* 1000000 ' — WiUJ - I — 100' U_ " 100? ° 99 qq ' 

T~0"0 yj 

Another illustration of an infinite geometrical series is found 
in the total distance traversed in passing from one point to 
another by passing first through -J- of the distance, then through 
half of the remaining distance, and successively in each sub- 
sequent movement through half of the distance which remains 
to the goal. One of the famous paradoxes of Zeno is to the 
effect that the hare cannot overtake the tortoise since the hare 
must first cover one half the distance intervening between their 
original positions, then one half of the remaining distance, 
then one half the remaining distance, and in each subsequent, 
interval of time one half of the distance which remains ; conse- 
quently the hare cannot overtake the tortoise as by this 
. . n _ 

A 8 Jf, 4 J/ 2 2 J/ 3 li/ 4 |5 

Hare Torloise 

process there is always distance intervening, and in the mean- 
time, further, the tortoise has advanced. If each of these 
intervals of space of the infinite series traversed required the 
same length of time, or any finite portion of time, the argu- 
ment would be sound, but you have here two infinite series 
each with a finite sum. As each space interval becomes 
smaller, the time required to traverse that space becomes 
smaller. The total distance evidently is the sum* of 8 -f- 4 + 2 
+ 1 + \ -h •• ■ which has the sum 16, both by the formula and 
by the figure ; to cover the whole distance requires 1 hr. and 
this takes the hare up to B. The fallacy in Zeno's argument 
lies in restricting the discussion to limited intervals of space 



178 UNIFIED MATHEMATICS 

and time preceding the instant and place at which the tortoise 
is overtaken. 



3. Geometrical means. — If a, x, b form a geometrical series 

- = -, whence x 2 = ab, x = Vab. 
x a 

If a, ar, ar 2 , ar 3 , ai A , • •• ar n ~ x , ar n , b form a geometrical series, 

evidently 

b = ar n • r = ar n+1 , 
i 

whence r =( — ) , and the series is 
W 

1 2 ^i w+1 

a - \ , a l - ) ot b. 



fb\ n+l fb\' +l /b\ n+l /b\ 



PROBLEMS 

1. Sum to twenty terms the series 2, 4, 8, 16, •••. 

In the following series give the sum to twenty terms, and 
where possible give the sum- "to infinity." 

2. 3, - 6, + 12, - 24, + 48, .... 

o Q 1 _1 1_ ... 

°* ^> 2' 1 2> 7 2> 

4. .1717171717 ••-, or .if (repeating decimal). 

5. .01717171717 -, or .Oif. 

6. 3.16161616 -.. 

7. 5, 15, 45, 135, •••. 

8. 4, 10, 28, 82, 244, ... the nth term being 1 + 3 n . 

9. 3, 7, 11, 15, 19, .... 

10. 3, — 7, — 17, - 27, .... 

11. v, v 2 , v z , v*, .... Sum to n terms. 

12. 1, r, r 2 , r 3 , r 4 , .... 

13. (1 + i), (1 + i) 2 , (1 + i') 3 , (1 + i)\ ... Sum to n terms. 

14. The number of direct ancestors which an individual has 
is represented by the series, 2, 4, 8, 16, — . Find the total num- 
ber in the preceding 10 and in the preceding 20 generations. 



GEOMETRICAL SERIES AND ANNUITIES 179 

15. According to Gal ton's law of heredity the parents con- 
tribute to the hereditary make-up of an individual i of what is 
contributed by all the ancestors ; the grandparents contribute 
J ; the great-grandparents, of whom there are eight, contribute 
i ; and so on ; hnd the total contribution. Find the individual 
contribution of a single individual four generations back. 

16. Insert three geometric means between 2 and 17 ; com- 
pute to one decimal place. 

17. Insert one, two, and three geometric means between 
1 and 2. 

18. One of the so-called " three famous problems of an- 
tiquity " is to construct, using only ruler and compass, a cube 
which is double the volume of a given cube. This problem was 
very soon reduced to the insertion of two geometric means 
between a and 2 a ; insert two geometric means between a 
and 2 a and show that this gives the algebraic value of the 
side of a cube of double the volume ; the geometrical solution 
has been demonstrated to be impossible if ruler and compass 
are the only instruments of construction. 

19. In the population statistics on page 65 find the population 
of the United States in 1810 and 1910 ; between these two 
numbers insert 9 geometric means ; find r ; this represents the 
approximate decennial rate of increase in the population at 
1810 which would give the final actual population in 1910. 
Compare with the/actual census figures at the end of each ten 
years. 

20. How would you find the regular annual rate of increase 
in the population of the United States, in other words the fixed 
annual percentage increase in population which would change 
the population from 7.2 millions in 1810 to 101.1 millions in 
1910 ? 

21. The plunger chamber of an air pump is approximately 
J- of the total air capacity ; at each stroke -JL of the air in the 
receiver is removed ; after 10, 15, and 20 strokes find the 



180 



UNIFIED MATHEMATICS 



proportionate amount of air remaining in the receiver ; approx- 
imately how many strokes must be made to remove 99% of 
the original air ? 

4. Graphical representation — Geometrical series. — On the 

straight line y — rx of slope angle a, with tan a == r, if x = a, 
ar, ar 2 , a?- 3 , — the successive abscissas represent the terms of 



i - ' i ■ ■ ir \ i r ' i i ! — 




u 




















p ^4 ' 




1 /r\ & ' 


^7--r — t 


. .. _ :yy-K/\\r\ 


T )r<> V |^ n ' 




i l\y & ' 1 






/"% ' / ' 1 


./I iUi ,' 1 1 




io/ «' _K r 1 


± CPf "mat I i 


/ / \\ /\ It 


^ /J / r ?5 


i" I/ \ ' i ^ 




^X i / 1 'J 




-H <<L ~ a Kf 1 Hjj 






>l <y . ' i 1 ! if 


>' / / I ' ' 


/ • 'fo j' ■ ' 1 ■ ■ I 1 




y i« A \ < / i IV,,: 


' •>> i i I 


v / > 


/>' Cl \ jrl ' ' 


^ a , ' \ i ■ I | v ' 




O \ /A AU \ A 3 \' ;A n ' \ \ ■ \\\ Q\\\\ \\lx 




1 / ' 1 ' 1 1 l 


y 


^4^- - 7 y ( ■ -Jrt k4_- . , _ . ^ 




y "t 




/ ' "i ■ ii 1 1 1 1 1 1 i 1 1 1 1 1 1 1 f 1 1 




' i : ' i Mi 


•1 1 1 ! 1 III 


'p ! 1 




-1 I ~ H- ~L - b" ' n : EE 


1 ■ I 


, . , . , :ii 1 



Graphical summation of the geometrical series, r < 1 

Note that the " sum to infinity,' 1 r < 1, is represented. 



the geometrical series ; by means of the 45° line through the 
origin these successive abscissas are readily constructed. 
However, a more favorable construction for a graphical treat- 



GEOMETRICAL SERIES AND ANNUITIES 181 

ment of the series regards each term of the series a, ar, ar 2 , 
a)- 3 , ••• as an addition to the preceding abscissa; the succes- 
sive additions to the ordinates, increments of the ordinates, 
will be ar 2 , a/- 3 , ar 4 , •••. By drawing the line, 



at an angle of 45° with the &-axis, the successive abscissas are 
readily constructed ; by drawing through the point Pi (a, ar) a 
line parallel to the o>axis it intersects the 45° line drawn 
through (a, 0) at a point M 2 such that 1\M 2 = P\A Y , since 
Z PiA x M 2 = 45° ; a parallel to the ?/-axis through M 2 intersects 
the line y = rx at a point P 2 such that M 2 P 2 = r * P\M« = ar2 > 
and A 2 P 2 = ar + ar 2 . Similarly, if P n represents the nth 
point found on our line, y = rx, 

M n P n = ar 11 , OA n = s n = a + ar + ar 2 + ••• ar n ~\ 

P A 

" n = tan a = r, whence P n A n = rs H ; A n M n = sr n — ar n ; A x A n 
OA n 

= s n — a ', but A 1 A n = A n M n , since the slope of the line 
A 1 M 2 M- 6 is 1. 

.-. s n — a = rs n — ar n , 

s n (l — r) = a — ar n , 
a — ar n 



s., = 



1-r 



If r < 1, as in our figure 1, a < 45°, and the two lines inter- 
sect at K to the right of the origin 5 and the points of intersec- 
tion M 2 , M 3 , - M n ••• will fall below K on the line BK. The 
abscissa of if represents the " sum to infinity" of our series, as 
it is evident that the series of triangles could be continued 
indefinitely in the opening OKA. Evidently also, solving 
y=x — a,y = rx, 



1-r 



182 



UNIFIED MATHEMATICS 





Vr- P / 




v~k 




-.. -j + _ _ z 






a r z 






2 s* c z: 




^ -^t ? -f- 






,/ t * 


2 .„. ,. J ;t 






J-n l,* 1 -A ; 


> ,« t 


J ,/ .4. . 


.,2 d X . 






&v • / t 


h ft 1 / h 


/ v / 


^ ! ItfV i 


Z _i_ liSG t . 


<<> ^ v' A I 


~CV „ "^ ' X 


4'Xy 41 < t 


t\'JL. *?-t -< I 


■'vy? ;jgt X fc 


^ - £ > + ?- 


I / '53 




■p ,-/ z!ct"'i£: _u;- idl' 


-t2/f- — -r-'TTjlA i 1*- 




/ t / \ f s» H 


I v J. 1 


/ 7TI _ ^ . _ 4. .. 


x- 1£t --/-- t 


/- S$ r t 


y iBi f 4t ■■ 


/ t Y £ f 




R-j ar r< r r t 


i>f 'IlQWi 


yit ytizfu X I 


Z^L\ J I T \ 




__ l ^Lj _, \ _J_ j_ . _ 


/ «t y c t t 


J- I ** I -t t 








L 1 " i*" ia.it ^."2 A3 Arv lit 




r _ ? ^__.lfc l_L_! rSrvptfc == 1 %- — -| 






:: :r-j^ : :: : : : -:: ::: : :: : ::-:[ :::::: ::: :: 












_i_ 1 



Graphical summation of the geometrical series, r > 1 

For r > 1, the figure is quite similar ; the two lines, y = rx 
and y = x— a, diverge ; tan a = r, a> 45°. 



Evidently, 

whence 
as before 



A n M n = A } A n , 

AP n = rs n , 

A n M n = rs n — ar n , 

OA n = s. 

A X A H = s-a, 

s n — a = rs n — ar n . 

s n (l — r) = a — ar n , 

a — ar n ar n 
s, = or 



1-r 



r-1 



GEOMETRICAL SERIES AND ANNUITIES 183 

5. Historical note. — Arithmetical and geometrical series 
are found in the oldest mathematical documents known, both 
in the remains of ancient Egypt and of ancient Babylon. The 
system of numbers used by the Babylonians as early as 2000- 
3000 B.C. was sexagesimal, increasing by powers of 60 in 
geometrical series. Further an early Babylonian clay tablet 
gives the portion of the moon's surface illuminated on each of 
fifteen successive nights from new moon to full moon by a 
geometric and an arithmetical series. The moon's surface is 
conceived as divided into 240 parts ; on the first five nights 
5, 10, 20, 40, and 80 parts, respectively, are illuminated and 
on the , following ten nights, 96, 112, 128, 144, ••• and on 
in arithmetical progression to 240. 

The Egyptian manual of mathematics of 1700 b.c. (or there- 
abouts), includes two rather complicated problems on arith- 
metical series, involving also the insertion of means, and one 
problem involving the summation of a geometric series. 

The equivalent of a general formula for summation of a 
geometrical series was first established rigorously by the 
Greeks, and appears in Euclid's Elements, Book IX, prop. 36. 
The first summation " to infinity " of a decreasing geometrical 
progression was effected by the great Archimedes (287-212 
b.c.) who employed the formula in finding the area of a seg- 
ment of a parabola. 

In a great part of the later development of mathematics 
such series have played a prominent role, in some measure 
because of their own intrinsic importance and in some measure 
as fundamental in the discussion of other types of series. 

6. Annuity formulas, a. Accumulated value of an annuity. — 
The geometrical series plays a large role in the theory of 
investments and insurance. We have shown (p. 54) that at 
rate i per year, compounded annually, 1 will amount in n years 
to (1+ i) n ; if 1 is invested at rate i at the end of each year, i.e. 
annually, for n years, these payments constitute an annuity of 
1 for n intervals, at i per annum. The total accumulated 



184 UNIFIED MATHEMATICS 

value 5^, at the end of n years of such an annuity, is the sum 
of the geometrical series 

(i + iy-\ (i + rp- 2 , (i + 0"-*, (i + i) n -\ - (i + Oi *> 

since the first payment made at the end of the first year ac- 
cumulates for (n — 1) years, the second for (n — 2) years, •••, 
and the n\h payment of 1 is made at the end of the n years. 
The sum, called the amount of the annuity, 

■ . _ (i+Q--i 

b n i — ; ) 

I 

represents the accumulated value of an annuity of 1 per inter- 
val for n years or intervals at a rate of i per year or interval. 

b. The annuity which will accumulate to 1. — Very evidently 
K per annum will produce at the end of n years 

Ks n = K • * — I ~~ ; the annuity which in n years will 
i 

amount to 1 is evidently the value of K which makes Ks n = 1 ; 
this value is — = 

c. Present value of an annuity of 1. — The accumulated 
value of 1 to be paid at the end of n years at i per year 
is (1 + i) n ; K will accumulate at the rate i in n years to 
K(l + i) n ; this means that K dollars (or units) in hand ac- 
cumulates to K(l -f i) n dollars, and that K(l + i) n dollars 
to be paid n years hence is worth K dollars now, money at 
rate i per year. Hence the present value of 1 to be paid n 
years hence is a value of K which makes K(l + i) n = 1, or 

K = = v n , wherein v = The present value of 

an annuity of 1 per annum for n years when money is worth 
i per annum is the sum of the geometrical series, 

v, v 2 , v z , v 4 , v 5 , r 6 , v 7 , — v n . 

The first term v is the present value of the first payment of 1 
which is to be made 1 year from date ; the second term is the 



GEOMETRICAL SERIES AND ANNUITIES 185 

present value of the second payment of 1 to be made in 2 
years ; — ; v n is the present value of the final payment of 1 to 
be made n years hence. 

V — v n+1 1 — v n 1 — v n 



tt,7i = 



1-v 1_ 1 



d. The annuity which 1 will purchase. — The present value of 

]_ r^n 

K per annum for n years is Ka-^ = K ; — ; the annuity 

% 

which is worth 1 at the present time is evidently a value K 

1 = i 
a^ l — v r 



which makes Ka» = 1, whence K= — = This is the 



annuity which 1 will purchase. 

e. Summary of interest functions. — 
These six functions, 

r n = (1 + i) n , accumulation of 1, 

v" = (1 -(- i)~ n i discount value, 

s^ = ^ f , accumulated annuity value, 



i 
_!-(!+ i)- n _1 -v 



V| (i + O re -i 



, present value of annuity, 



annuity to accumulate to 1, 



1 i 

and — == , the annuity which 1 will purchase, 

afci l-(l + 0~ n 

are of fundamental importance in the valuation of bonds, and 
in all problems where stipulated payments are to be made at 
stipulated intervals, and also in the theory of interest. 

If interest is to be compounded semiannually and pay- 
ments are made semi annually, the interval can be considered 
•is 6 months and the rate of interest as one half the stated 
rate ; similarly the interval can be considered as 3 months and 
the rate of interest as one fourth the stated interest if interest 



186 UNIFIED MATHEMATICS 

is compounded quarterly and payments made quarterly. 
Other types of problems with payments falling between in- 
terest periods are beyond the scope of this work. 

PROBLEMS 

1. Compute the value of s for 6%, 5%, 4%, using loga- 
rithms to obtain (1 + .06) 20 , (1 + .05) 20 , and (1 + .04) 20 . What 
percentage of error is introduced using four-place tables ? Dis- 
cuss the effect in finding the accumulated value of an annuity 
of $ 100. per year for 20 years. Check by the tables given at 
the back of this book. 

2. Compute a—,, and discuss as in 1. 

3. Find by logarithms from your values of s W[ and a~, the 
annuity which will accumulate to 1 in 20 years, and the 
annuity which 1 in hand will purchase. 

4. If payments of $50 per year are made semiannually 
and interest is compounded semiannually, find the accumu- 
lated value of this annuity at the end of 20 years for a 
nominal interest rate of 6 % , 5 % , and 4 % respectively, per 
annum (3 %, 2-j- %, and 2 % per interval). Use^the tables. 

5. What annual payments continued for 10 years are equal 
to $ 1000 cash in hand ? 

6. What annual payments continued for 10 years are equiv- 
alent to $ 1000 to be paid at the end of 10 years? to $ 1000 to 
be paid at the end of 20 years ? 

7. Prove s^ = (1 -j- i) n a^-\ ; and a^ = v n • s^. Discuss. 

8. Show algebraically that = i. 

Note. — The difference between 1 in hand and 1 to be paid in n years 
is simply the earning power of the 1 in hand for this period of n years ; 
if money is worth 6 °fo per year, 1 in hand will earn every year for n 
years. 06 in addition to preserving itself ; 1 in n years is worth simply 1 
then : - — is the annuity for n years which 1 in hand purchases, and — is 



GEOMETRICAL SERIES AND ANNUITIES 187 

the annuity equivalent to 1 to be paid in n years ; the difference is the 
annual earning of the 1 in hand, i. 

9. Give the arithmetical series represented by the ordinates 
of y = 3 x -f- 7, for integral values of x from to 10. What is 
the sum of this series ? Between 1 and 2 interpolate 9 values, 
at equal intervals, and state corresponding series ; what corre- 
sponds to the tabular difference ? 

10. Discuss as in problem 9 the corresponding ordinates of 
2ij = -3x + 7. 

11. Sum to 20 terms the series 7, 5, 3, 1, — 1, •••. 

12. Write the 20th term of 7, 5, 3, 1, — 1, -. 

13. Write the tenth term of 7, 5, if-, ^f-, 

14. Find the arithmetical and the geometrical means be- 
tween 7 and 5. 

15. Historical problem. — It is related that an Indian 
prince who wished to reward the inventor of the game of 
chess suggested to the inventor that he should name the 
reward he desired. The scholar replied that he would take 
1 grain of wheat for the first square of the board, 2 for 
the second, 4 for the third, 8 for the fourth, 16 for the fifth, 
and so on in geometrical progression to cover the 64 squares. 
The prince agreed but found, on the computation, that the 
value exceeded that of his realm. Taking 10,000 grains as 
approximately a pint, make a rough calculation of the amount 
involved. 

16. Historical problem. — In textbooks of the sixteenth 
century the following problem frequently appears. A black- 
smith being asked his price for shoeing a horse replied that 
for the first nail he would charge one fourth of one cent (use 
this in place of farthing, or pfennig), i cent for the second 
nail, 1 cent for the third, 2 for the fourth, and so on for the 
thirty-two nails. Compute the price. 



188 UNIFIED MATHEMATICS 

7. Annuity applications. — Brief tables of the annuity func- 
tions are given at the back of this book ; somewhat larger 
tables will be found in the Bulletin No. 136 of the U. S. 
Bureau of Agriculture, which includes also a more extensive 
treatment of the subject of bonds and annuities by Professor 
James W. Glover. 

a. Common annuity. — To find the purchase price of an 
annuity of k dollars per interval for n intervals when the current 
rate is i per interval is obviously a direct application of the 
a^| table, as the purchase price is simply the present value of 
the series of payments. If the first payment of the annuity is 
to be made r intervals hence, a deferred annuity for n intervals, 
the price may be considered as the difference between an 
annuity for r — 1 intervals and an annuity for n + r —1 inter- 
vals, a^^j-, gives a payment every interval for n + r — 1 
intervals, the first made at one interval from the present time ; 
a f-r\ gives a payment every interval for (r — 1) intervals, the 
first as before, and the last (r — 1) intervals from the present 
time ; the difference is the value of the deferred annuity of n 
payments, first payment to be made at the end of r intervals. 

Some large banks, trust companies, and insurance companies 
do this type of business. Frequently a purchaser desires an 
annuity to be paid annually terminating with the death of the 
purchaser ; this involves then a life contingency, and the dis- 
cussion and solution of this problem require new methods and 
new tables. 

b. Farm loans. — To extinguish or amortize a debt by n 
annual payments of fixed amount is the type of problem which 
arises under the recent Farm Loan Act. Thus a farmer 
borrowing $ 10,000 at the bank at 5 % interest may desire to 
make such a payment as to extinguish the debt in 30 years, 
money being worth 5 % annually. The problem may be 
solved by considering the annuity which $ 10,000 will purchase 
at 5 % interest for 30 years or $ 10,000 X .06505, giving 
$ 650.50. The problem may also be solved by considering the 
interest as paid each year, $ 500, in addition to which an 



GEOMETRICAL SERIES AND ANNUITIES 189 

annual payment must be made to accumulate at 5 % to $ 10,000 
at the end of 30 years. This annual payment is found to be 
S 150.50. The value found, $ 650.50, maybe checked, as below, 
by using the s^ table. Or another check is to find the value 
at the end of 30 years of the $ 10,000 or S 10,000 (1 + -05) 30 and 
compare this with the value of S 650.50 x %> 

Suppose, on the other hand, that the borrower desires to 
pay approximately $ 600 per year, applying the extra amount 

each year to the debt. The annuity which 1 will purchase, — , 

is the function involved. The question here may then be put, 
For what period of years at 5 % interest will $ 10,000 purchase 
an annuity of $ 600 per annum ? We will consider only ap- 
proximate solutions, taken from the tables. 

The tables show that at 5 % interest $ 10,000 will purchase 

an annuity of $ 10,000 X — = % 10,000 x .06043 for 36 years, 

or $ 604.30 annually for 36 years ; S 10,000 will purchase an 
annuity of $ 598.40 for 37 years. 36 years would be taken, 
and this period of 36 years is provided for, as an amortization 
term, by the government. By paying every year $ 100 more 
than the interest, at the end of 36 years the accumulated 
value of this annuity, the excess $ 100 over the interest, would 
be worth at 5 % : $ 100 x s^] = S 100 x 95.8363 = $ 9,583.63, 
leaving $ 416.37 due at the end of 36 years. This amount 
with interest at 5 % should be the next and final payment. 

c. Sinking funds. — If a city issues bonds to be redeemed 20 
or 30 or 40 or n years hence, it is commonly desirable to pro- 
vide for the repayment of the bonds by an annual (interval) 
payment allowed to accumulate at i per annum (per interval). 
Similarly in business a manufacturing concern using an ex- 
pensive piece of machinery which has a probable lifetime of 
20, 30, or 40 years must provide for the eventual replacement 
of this machine by an annual payment, out of earnings, into a 
sinking fund. In this type of problem the function involved 
is the annuity which will accumulate to 1 in n years. Thus 



190 UNIFIED MATHEMATICS 

to provide for replacement of a $ 10,000 piece of machinery 
in 30 years money at 5 % requires an annual payment of 

$ 10,000 x — or $ 10,000 x l which equals $ 150.50 

Saol (1 + 30 - 1 

per annum. 

d. Bonds at premium and discount. — If a city issues bonds 
at 5 % when money is worth in the money market 4 %, the 
bonds will sell at higher than face value, since they pay on 
each $ 100 an annuity of $ 5.00 per year for the term of the 
bond, when investors are demanding only $ 4.00 per annum 
with security of capital. This higher price is, on the basis of 
money at 4 %, the present value Of an annuity of $ 1.00 per 
annum for the term of the bond or 1 x <%, at 4 % . The dif- 
ference between the par value or face value of a bond and the 
price offered by investors is called the premium (or discount, 
when the price offered is less) on the bond. 

If a city issues bonds at 5 % when investors are demanding 
6 %, a bond for $ 100 will sell at a discount of 1 X <Vi at 6 %, 
since the investor receives from these bonds not $ 6.00 per 
annum but only $ 5.00 per annum. Evidently the longer the 
bond has to run the greater would be the discount. 

In general terms the premium on a bond oi face value G, 
paying a dividend rate g, bought to yield j per annum is 

P = C(g — j)ajq at j per annum. 



PROBLEMS 

1. If a father sets aside annually $ 100 per year as a fund 
for his son when the latter becomes of age, to what will the 
fund amount at the end of 21 years, the money accumulating 
at 4 % interest ? 

2. If a farm mortgage of $10,000 draws 5 % interest and 
the farmer pays annually $ 600, what is the accumulated value 
at the end of 21 years of the excess payments of $ 1 00 per 
annum, accumulated at 5 % ? 



GEOMETRICAL SERIES AND ANNUITIES 191 

3. What annual payment will accumulate at 5 % in 30 years 
to $ 10,000 ? What annual payment would have to be made 
on a §10,000 mortgage, to extinguish the debt in 30 years, 
money worth 6 % ? 

4. Find in the tables the annuity for 30 years which 
$ 10,000 will purchase. 

5. What semiannual payment will accumulate in 30 
years (60 payments) to $ 10,000, interest being 5 % com- 
pounded semiannually ? 

6. Find the cost of an annuity of $ 10,000 per year to run 
for 10 years, 20 years, and 30 years, respectively, money being 
worth 4 <f . 

7. If a city issues $10,000 in bonds, what amount must 
be set aside annually to accumulate at 4 % interest to redeem 
the bonds at the end of 20 years ? 

8. Find the cost of an annuity of $ 500 per amiurn for 10 
years, the first payment to be made 10 years hence, 20 years 
hence, and 30 years hence, respectively. 

9. What premium can you afford to pay on a $ 10,000 
bond drawing 5 % to run 20 years, if money is worth 4 °/ ? 
What discount should you receive if money is worth 6 % ? 

10. What is the present value of $10,000 to be paid 20 
years hence, money at 5 % ? 

11. Which is the better offer for a piece of property, money 
being worth 5 %, a rental of $600 per year for 20 years, or a 
price of $10,000? A rental of $600 per year for 10 years, 
and $700 per year for the following 10 years, or $12,500? 
Assume no change in the price of the real estate in 20 years. 

12. What sum at 4 % interest should a railroad set aside 
each year to replace engines worth $35,000, which have an 
estimated life of 25 years? to replace buildings worth 
$ 1,090,000 which have an estimated life of 100 years ? 



192 UNIFIED MATHEMATICS 

13. If a man invests $ 100 each year for 20 years, what 
annuity, for 20 years, can he purchase at the end of the first 
20 years, money at 5 % interest ? 

14. If a man agrees to take $ 1000 a year for five years for 
a house originally offered at $ 5000, what is the discount when 
money is worth 5 % ? 

15. At 5 % interest what annual payment for five years is 
equivalent to $ 5000 cash in hand ? 

16. Answer questions 14 and 15, assuming that the first 
$ 1000 is to be paid immediately. 



CHAPTER XII 

BINOMIAL SERIES AND APPLICATIONS 

1. Binomial series. — The expressions (1 + z) 4 , (1 -f i) 20 , ... 
can be developed in powers of i by means of the binomial 
expansion, 

(a + »)" = a n H a" l aj + -^- — — l a n 2 .t 2 + -*— — ^ v ' • a n " 3 ^ 

m(k - 1)Q - 2) ... to (r - 1) factors _ a n- r+v -i , ... 
1 .2 . 3 • 4 .- (r - 1) T ' 

In particular, if a = 1, 

(l + *)■ = i + \- x + '-i^D . * + *(" -1X" 3 -' 2 ) • ^» 

n(n — l)(n — 2)(n — 3) to (r — 1) factors r _ l 

1.2 : 3.4.« (r-1) "* + " % 

This formula expresses a rule for the formation of successive 
terms of the expansion of (a + x) n or (1 + x) n ; the first term 
contains a with the exponent ?i of the binomial ; the second 
term has as coefficient the exponent, or index, of the binomial, 
x appears to the first power, and the exponent of a decreases 
by 1 ; the coefficient of the third term has two factors in 
numerator and denominator, in the numerator n(n — 1) and 
in the denominator 1 • 2 ; a appears with exponent 1 less 
than in the preceding term, and x with exponent 1 greater ; 
each following term can be obtained from the preceding by 
introducing one further factor in numerator and denominator 
and at the same time decreasing the power of a by one and 
increasing that of x by one ; the further factor in the new 
numerator is one less than the last one introduced there, and 

193 



194 UNIFIED MATHEMATICS 

the further factor in the new denominator is one greater than 
the last one of the preceding denominator ; the coefficient of 
the term in x n ^ contains in the numerator (r— 1) integral factors 
from n down, and in the denominator (r — 1) integral factors 
from 1 up; a; appears with exponent r — 1 and a with the 
exponent which added to r — 1 makes n, i.e. n — r + 1. 

Illustrations of the binomial expansion. 

a. (a + x) 3 — a 3 4- - a 2 x -f — ax 1 A — — X s 

K J 1 1-2 1 -2-3 



= a 3 -f 3 a 2 x + 3ax 2 + &• 

1st 2d 3d 

b. (a + x) lb = a 15 + — a u x + *5jJi a n x* + 

J. JL • A 

a & x 9 



10th Term 
15.14.13-12.11.10.9.8.7 



1.2.3.4.5.6.7.8.9 

Note that it is well in writing the tenth term to begin with 
x 9 ; then a enters to the sixth power as in every term of this 
expansion the exponents of a and x together make 15 ; the 
denominator contains 9 factors, 1, 2, 3, • • • 9 ; the numerator 
contains 9 factors, which should be counted as they are writ- 
ten ; finally cancellation should be made, giving 

5 • 7 • 11 • 13 a«x\ 
c. ( a -a) 15 = a 15 - — a 14 x +^^ aV- 15 ' 14 ' 13 a x V 

10th Term 

+ 5 . 7 • 11 • 13 aV + -. 

a. (i + *)" = i + 15, +^x* + 15 ; 1 9 4 ; 13 a ? + 

10th 

15-14. 13-12^ + ... 5 . 7 . 11 , 13 , 9 , 
1.2.3.4 

Note that the powers of a in each term can be dropped, as 
every power of one equals one. 



BINOMIAL SERIES AND APPLICATIONS 195 

e. Compute to 4 decimal places (1 + .04) 15 . 
(1 + -04) 15 = 1 + 15(.04) + l^l 4 (.04)2 + lg 1 ' t 1 2 4 ' 13 (.04)* + 

1^14-13.12 15.14.13.12.11 ^ 

1-2-3-4 ^ ; 1.2.3.4.5 V ' ; ' 

.60 second .0291 1.00000 first term 

7 .12 .60000 second term 



4.20 5 ).00349 fifth .16800 third term 

.04 .00070 .02912 fourth term 



3 ). 1680 third term 11 .00349 fifth term 

.0560 .0077 .00031 sixth term 

.04 .04 1.80092 Ans. 



.002240 .00031 sixth 

.02240 10 times 
672 3 times 
.02912 fourth term, to be multiplied by .12 

Note here the method of computation given at the left ; each 
term is obtained from the preceding term ; three new factors 
of which one is .04 enter into each succeeding term, two in the 
numerator and one in the denominator ; these three factors after 

the second term (.60) are U X - 04 or 7 x .04 ; then 13 X - 04 ; 

Z o 

then 12 x - 04 or .12; then 41 x - 04 , which might well be 
4 5 

treated as .008 x 11 ; then the factor : — would give the 

6 

seventh term from the sixth, making about 2 in the fifth deci- 
mal place. 

The expansion of 

(l + g )» = l + n g + w ^- 1 > a » + n(w ~ 1)(n ~ 2) ^+ - 
1.2 1-2-3 

may be written as follows : 

+ (n-4), ; 



196 UNIFIED MATHEMATICS 

in which T 2 , T 3 , T 4 , T b} ••• designate the second, third, fourth, •» 
terms respectively. This type of representation, in which 
each term is obtained from the preceding, is frequently of use 
in statistical work and in computation. 

/. Write the sixth and sixteenth terms of (1 + x) 17 „ 

6th Term 16th Teem 

17.16.15.14.13 . 17-16... 15 

1.2.3.4.5 ' 1-2... *' 

g. What decimal place is affected by the sixth term of 
(1.06) 17 ? 

17 • 16 • 15 • 14 • 13 



1.2.3.4.5 



(.06) 5 ; (.06)3= .000216 ; (.06) 4 = .00001296 



(.06) 5 = .00000078- ; multiply ,00000078 by 4, this by 7, this 
by 13, and then by 17, rejecting any beyond 3 significant fig- 
ures ; this gives .00000312, .0000218, .000286, and finally .00476. 
Note that the computation of six terms of (1 + .06) 17 involves not 
very much more numerical labor than this determination. 

h. Write six terms of (a — 3 a?) 12 . 

aP - 12 aP (3 x) + — - o">(3 xY - 12 ' n ' 1Cf a 9 (3 xY 
1-2 J 1-2- 3 

12.1L1Q.9 (3 )A 12-11.10.9.8 7(3 v 

+ 1.2.3.4 a{3x) ~ 1.2.3-4.5 a{Sxy 

a i2 _ 36 a n x + 2 • 3 3 • 11 a w x* - 2 2 . 3 3 • 5 • 11 a¥ -f 5 • 3 6 • 11 a¥ 
-2 8 -3 7 .llaV. 

It is not necessary or desirable to perform the multiplica- 
tion in such an expression as 2 2 • 3 3 • 5 • 11 ; such terms, if 
desired numerically, are usually obtained progressively from 
preceding terms as in example (e) above. 

PROBLEMS 

1. Expand to 6 terms, (a + x) & , (a + x) u , (a + 2 a) 10 ' 
(a-r-3xy. 



BINOMIAL SERIES AND APPLICATIONS 197 

2. Write 5 terms in simplest form (prime factors) of 
(1-fz) 6 , (1+*) 14 , (1 + 2 a) 10 , (l-3.r) 9 . 

3. Compute to 4 decimal places (1 -f .05) 6 , (1 + .05) 14 , 
(1 + -05) 10 , (1 - .05) 9 . 

4. Compute to 2 decimal places the value at the end of 
10 years of $ 100 placed at interest at 6 % compounded annu- 
ally ; use 100 x (1 + .06) 10 . How could you use the result 
obtained to find the value at the end of 20 years ? 

5. From problem 3 give the amount at the end of 6, 14, 
and 10 years, respectively, of $ 256 at 5 % interest, com- 
pounded annually. 

6. Find the value at the end of 6, 14, and 10 years respec- 
tively of an annuity of 1 per annum, paid at the end of each 

year, interest at 5 % . Use the formula s^ = v ;- ~ and 
the preceding results. 

7. Compute the values in 3, 4, and 5 by logarithms and 
compare. 

8. Given 2 10 = 1024, find to 1 decimal place (2 + .01) 10 . 
Ans. 1076.4. 

Find also (2.1) 10 to one decimal place, and check by logs. 

9. Find (12.3) 3 to five significant figures. 

10. Find the amount at the end of 20 years of $ 100 placed 
at interest, 3%, compounded semiannually. 

11. Write the 8th term of (1 - 3 a) 17 and of (1 -f- xf\ 

12. How many terms in (1—3 x) 11 ? Write the middle 
;erms. 

13. Write in simplest form the coefficient of x 6 in (1 — 2 x) 25 . 

14. Write the series for (1 -+- sc) 5 , (1 + a;) 6 , (1 + x) 7 , and 
(1 -f x)\ What is the sum of the coefficients ? 

(Note. — Substitute 1 for x.) 

15. What is the sum of the coefficients in (1 + x) 19 ? 

16. Write 7 terms of (1 + Va;) 10 and of (1 + -tyx)™. 



198 UNIFIED MATHEMATICS 

2. Proof of (a + x) n = a n + J a"- 1 * + W( \" ~ ^ 0-2*2 ^ 

1 1 • <© 

>'th Term to a Total of (f — I) Factors 

ji(n-l)(fi-2) 3 - . w(n-l)(n-2) - +1 x 

12 3 ^ 1-2 34 (r-1) "*" ' 

The proof of this expansion for positive integral values of n 
is effected by the process called mathematical induction. 

Evidently ' (a + x) 2 = a 2 + 2 ace + a? 2 , follows the rule ; 
also (a + a;) 3 = a 3 + 3 a 2 x + 3 ax 2 +. x*, follows the rule. 

By the rule, 

, , ^4 4,4 o ,4.3 02l 4-3.2 ,. 4-3.2-1. 

= a 4 + 4 a 3 aj + 6a 2 x 2 + 4 aa 3 + x" ; 

by actual multiplication we find the same series, showing that 
the rule as given holds for n = 4. 

Assume 

/ , \« „ *-i ,n(n — 1) „_„ . 

(a + a;) 71 = a ra + na n ^ H — * — - — '- a n 2 x 2 + ... 

n(n - l)(n - 2) ... (n - r.+ 3 ) +2 1._ 2 
1.2.3-.(r-2) "~ a X 

r*Tii Term 

4. n(» - l)(n - 2) - (n - r + 2) .. n - r+ , x r-i . 
+ 1.2.3.,(r-l) " ' 

multiply by a + x — a + x 

(a + x) n+1 = a" +1 + (ti + l)a w -^ +(« + ^ ~ J^ W 2 + '" 

The New ^th Term 

Aip — l)(n - 2) - Q - r + 3)Q - r + 2) 
V 1- 23 ••• (r-2)(r-l) 

n(n - l)(n - 2) ... (n - r + 3) V w -^ r -i 
1-2- 3 ...(■>• -2) J 



BINOMIAL SERIES AND APPLICATIONS 199 

Note that these two coefficients of the new rth term have 
the first (r — 2) factors of numerator and denominator the 
same ; multiply the denominator and numerator of the second 
term by r — 1 and then add the numerators, taking out the 
common factors, giving 

New ma Teem Common Remaining Remaining 

Factobs Factor of Factor of 

First Term Second Term 

l [„(„-!) - (»-r+3)] [(»- r+2) + (r - 1)] \ -.-^^ 
i 1.2.8.4.6...(r-l) I ' 

The new rth term may be written, then, 

(n + l)(n)(» - l)(n - 2) - Q - r + 3) ( „ +1) _ r+ , ,. 
1.2-3.4... (r-1) ' 

the rth term of (a + x) n+1 is formed according to the rule with 
(n -f- 1) substituted throughout for n ; the numerator contains 
(r — 1) factors beginning with (n -+- 1), and the denominator 
contains the same number of factors beginning with 1. 

Hence if this expansion assumed for (a -f- x) n is correct for 
any value n, it is correct for a value one greater, n-j-1. The 
theorem is true, by trial, for n = 4 ; hence it is true for n = 5 ; 
since it is true for n = 5 it is also true for n = 6 ; ■•• and so 
for every integral value of n. 

3. Binomial series ; any exponent. — The equation, 

(i + a? )» = l + ^ .+ Z^zl) X 2+ n(n-l)(n-2) 
1-1.-2 1 • 2 • 3 

■ n(n-l)(n-2)(n-3) . ... 
1.2-3-4 

can be shown by methods of the higher mathematics to hold 
for all values of n when — 1 < x < 1. This means that a 
series for (1 + x)* can be obtained by substituting n = \ in 
the above formula. Similarly (l-fz) -3 and (1 + x)~ 7 , (1 + x) - ^ 
or (1 + x)^ 2 can be developed in powers of x, when \x\ < 1, by 



200 UNIFIED MATHEMATICS 

substituting for n the values — 3, and — 7, and the like in the 
above formula. These series are of frequent use in statistical 
work. Thus if money is worth 6 °/ per annum the interest 
for one half year is not 3 °j of 1, since this rate continued for 
the full year would give at the end of one year in addition to 
the 6 % of 1, the interest on the 3 % of 1 for one half year ; 
the interest on 1 for one half year is taken to be 

/=(l + .06)*-l, 

and this rate of interest per half year accumulates . at the end 
of two half years a principal of one to (1.06), or is equivalent 
to 6 % per year. Similarly the effective rate of 6 % per 
annum means that the interest for one fourth of a year will 
not be .015 times the principal, but rather (1 -f .06)* — 1, 
since this is the rate of interest for one quarter of a year 
which continued for four quarters will accumulate a principal 
of 1 to 1.06, since [(1 + .06)*] 4 = 1.06. 

In our illustrative problems we will assume that the 
terms that follow any given term in the expansion of an ex- 
pression like (1 + a;) 5 are together less than the last term 
given ; the general proof of the convergence of these series 
is reserved for the calculus ; however, it is evident here that 
each new factor, when x is less than 1, diminishes in value and 
finally the terms are in turn respectively less than the terms 
of a geometrical series with ratio x ; thus, below in the ex- 
pansion of (1 + .06) 1 , beyond any given term the terms are 
respectively less, term by term, than the terms of a geometri- 
cal series with ratio .06; the sum of all terms beyond the 

fourth term is certainly less than — , wherein a is the fourth 
J .94' 

term, since the corresponding geometrical series, even to an 

infinite number of terms has only this sum, — — — = 

J ' 1 - r .94 

Note particularly that the expansion of (1 -j- x) n for values 
of n other than positive integers leads to a series which has 
no termination, i.e. to an infinite series ; this series is valid 



BINOMIAL SERIES AND APPLICATIONS 



201 



and has meaning only when |rcj<l; similarly any infinite 
series given by (a -4- x) n has meaning only when I— < 1. In 

our further discussion this limitation will be consistently 
assumed. 

As au illustration of the possible absurdity from the point 
of view of finite series of the infinite series given by (1 + x) n , 

let us take the fraction which may be written (1 — a?) -1 , 

1 — x 

and developed by the binomial theorem as, 

1 + x -j- x 2 + a 3 + x 4 + x 5 -f -.. 
For values of x numerically less than unity this series is valid, 
but if you put x equal to 3, or — 5, or other value greater than 
unity, you obtain an absurdity. 

4. Illustrative problems. — a. Compute (1 + .06)2 to 5 deci- 
mal places. 



(l+x) 



x + 



Ki-D 



z 2 + 



Ki-i)q-2) _ x3 + _ 



1-2 1.2-3 

(1 + .06)* = l + i(.06) - 1 (.06)2 + T i_ ( .06)3 + ... 

= 1 + .03 - .00045 + .0000135 + ... 

= 1.02956. 
i 
If the value of (1.06) 2 were desired to eight decimal places, progressive 

computation would be desirable, appearing as follows : 



T 1 = l 

T 2 = K-06) Ti = .03 

rr l/.06\™ 

T3 = -2(^j T2 = - 



.00045 



r 4 = 



r 6 = 



TV 



.06 



+ .0000135 



.06 



r 4 =-. 000000506 



.06™ 



+ .000000021 



4 ) .0000135 
.000003375 

5 

.000016875 
.03 
.00000050625 
.042 
1012 
2024 
.00000002125 

1.030013521 

- .000450507 

1.029563014 



.06 



r 6 = - .ooooooooi 



202 UNIFIED MATHEMATICS 

The letters Ti, T 2 , T 3 , T 4 , ••■ represent the first, second, third, fourth, 
and succeeding terms ; each term is obtained from the preceding, intro- 
ducing the new factors. 

b. Write six terms of the series for (1 + x)* and for (1 — x) z . 

1-1 1-1-3 1-1-3-5 

i 1 2*2 2*2*2 2* 222 

1-1-3 -5-7 

2' 2 ' 2 ' 2 ' 2 

+ 1.2.3.4.5 X5+ - 

= l + laj-laj-. T 2 -lx- T z -\x- T A -&x- T 5 . 

Alternate terms after the second are positive and negative. 

1 / -1\ 1-1-3 1-1-3-5 

i 1 2 \ 2 / 2 2 2 2 2 2 2 



1-1-3 -5 -7 
2*2* 222 



x 5 



1-2-3.4.5 
= 1 — \x- \x* - j\x* - T | ? x 4 

= i-ix + |x. r 2 + ix. r 3 + fx- r 4 + T Vx- t 5 + .... 

Every term after the first is negative. 

c. Compute V.98 to 5 places. 

(1 - .02)* = 1 - £(.02) + K-i) (.02)2 _ K- i)(- I) (-.02)3 + ... 

= l _ .01 - .00005 - .0000005 - ••• 
= .9899495 or .98995 to 5 places. . 

d. Compute the cube root of (1012) to 6 significant figures. 
(1012)* = (1000) 3 (1 + .012)* = 10(1 + .012.)* 

(1 + .012)* = 1 + K- 012 ) + ^-* 1 ) (.012)2 + Ki" 1 )^- 2 ) (.012)3 + ... 
1-2 1 • 2 • o 

= 1 4. .004 - .000016 + .0000001 - ... 

= 1.003984. 

(1012)*= 10 x 1.003984 = 10.03984. 

e. Compute (1.05)* to 4 places; treat this as 1.05 x (1.05) 2 ; 
.05 may be taken as ^V- Check by logarithms. 



BINOMIAL SERIES AND APPLICATIONS 203 

/. Compute the square roots of 26 and 30 to 4 places. 

\/26 = 26^ = (25 + 1) 2 = 5(1 + ft)* = 5(1 + .04)* 

= 5(1 + .02 - .0002 + .000004) 
= 5(1.019804)= 5.09902. 
30* =(25 + 5) 2 = 5(1 + .2)* = 5(1 + .1 - .005 + .0005 - .0000625) 
= 5(1.0954375) =5.47718. 
Check roughly, using logarithms. 

g. Compute (1.05)~ 7 to 5 significant figures. 

(1.05)-? = 1 - 7(.05)+ - 7 ^- 8 ^ (.05)2 + ~ 7( ~| X 3 ~ 9) (.05)3 

-7(-8)(-9)(-10) ( m 4 -7(-8)(-9)(-10)(-ll) 

1 .2.3-4 1-2- 3- 4- 5 v " J 

- 1 _ .35 4. .07 - .0105 + .0013125 - .0001444 + .0000144 = .71068. 

5. Historical note. — The binomial theorem as applied to 
(a + b) 2 and (a -}- bf was well known to Euclid (320 b.c.) and 
other early Greek mathematicians. The great Arabic mathe- 
matician and poet Omar al Khayyam (died 1123 a.d.) extended 
the rule to other positive integers. In China there appeared 
in 1303 a work containing the binomial coefficients arranged 
in triangnlar form, the so-called Pascal (1623-1662) triangle 
of coefficients. 

111111 
1 12 3 4 5 6 

11 1 3 6 10 15 21 

12 1 1 4 10 20 35 

13 3 1 1 5 15 35 

14641 16 21 
1 5 10 10 5 1 17 
1 6 15 20 15 6 1 1 

Pascal's Triangle Chinese Form of Tri- Stifel's (1486-1567) 
of Coefficients, angle of Coeffi- Triangle of Coeffi- 

Printed 1665. cients. cients, 1544. 

The general rule for any exponent — was first discovered 

n 
by Sir Isaac Newton, and made known in a letter of date Oct. 



1 


1 














7 




1 2 
1 3 


1 

3 


1 












1 4 


6 


4 


1 










1 5 


10 


10 


5 


1 








1 6 15 


20 


15 


6 


1 



204 UNIFIED MATHEMATICS 

24, 1676, to a friend named Oldenburg. It is of interest to 
note that Newton wrote each coefficient in terms of the coeffi- 
cient immediately preceding, following the lines indicated in 
our numerical problems above. The complete proof for the 
general case, any real or complex imaginary number, was finally 
effected by a brilliant Norwegian mathematician, Abel (1802- 
1829), in 1826. 

PROBLEMS 

1. Find the coefficients of the first five terms of (1 + xy 
and (1 — x)t and use them in the next two problems. 

2. Compute to 3 significant figures the square roots of 27, 
28, and 29 as 5(1 + .08)*, 5(1 + .12)*, and 5(1 + .16)*, respec- 
tively. 

3. Compute the square roots of the first eleven integers, 
to 5 places, taking V2 as ^ (196 + 4)* = ff (1 + ±\)\ V3 as 
2(1-1)2, V5 as 2(1 + i)*, V6 as (4 + 2)* = 2(1 + .5)*, 
V7 as 1(25 + 3)*, V8 as 2V2, VlT as J (100 - 1)*. Check 
the first four significant figures by logarithms. 

4. Write 5 terms of (1 -+- a;)*, and of (1 — as)*. 

5. Use these to compute the values of V7 and v^9, as 
2(1-*)* and 2(1 + *)*. 

6. From the cube root of 9 find the cube root of 3. 

7. Find the cube root of 6. 

8. Using the cube roots of 6 and 3, find the cube root of 2. 

9. Compute (1 + .05)* to 5 places. 

10. Write 5 terms of (2 - 3 as)~* 

11. Write the 6th term of (2 - 3 a;)" 7 . 

12. Find the value to 4 significant figures of — - ^y 



expanding (1 — .02) 



i Vl - .02 



BINOMIAL SERIES AND APPLICATIONS 205 

13. Expand in powers of i to 4 terms and What 

1-H' 1 — £ 

is the approximate percentage error in using (1 — i) and 

(1 -f i) as multipliers, respectively, instead of : and — — ., 

when i = .01, .05, .005, and .5, respectively ? 

14. Eind the fifth root of 35 correct to 2 decimal places. 
What is the shortest way? Compute the root to 5 places. 
What method can you use ? 

15. Write the term containing x e in the expansion of 
(1-3 a:)-* 

16. Write the first five terms of (10 -f- .3) 8 ; of (10.3) 6 ; of 
(10.3) -6 and give the value to 4 significant figures. 

17. Time yourself on writing and simplifying 10 terms of 

12 1 

the following five expansions : (1 -f x)*, (1 — x) *, (1 — 2 x)~*, 

18. Compute correctly to 4 significant figures, using the 

formulas for and Time yourself on the exercise, 

1 + i 1 — i 

18 _27_ J>£ 62 J>8_ 

.98' 1.02' 1.03' .99' 1.05* 

19. Compute the following to three significant figures, tim- 
ing yourself : 

(1.06)*, (1.06)T2 ? (1.05)*, (1.06)3, (1.10)* (1.06)"*. 



CHAPTER XIII 



RIGHT TRIANGLES 

1. Right triangles. — To apply our trigonometric work to the 
numerical solution of right triangles place the triangle under 
consideration in quadrant I in proper position to be able to read 
the trigonometric functions of one acute angle. 

rsin p 



jz zcrr 


i-R 


_ _ __ __ _ -±:£ 


_£ 


„--£ -Q 










-»' V' 




^~*" yot x 






A T T -t,.' 



x = r cos a 

y = r sin a = r cos (3 

= cot p 



y_ 

x 



Fundamental formulas of the right 
triangle 



cot a = - = tan B 

y 

: 2 + y 2 = r- 

90°.^ 



cos a = - 
r 



or 



a + p 

The equation x = r cos a may be written 

r = — - — , as occasion demands ; similar transformations are to 
cos a 

be effected upon the other equations given. 

Given a with x, y, or r ; or /? with x, y, or r ; or two of the 
lengths ; these formulas enable us to solve the right triangle 
completely for the remaining three parts. 

The student is advised to draw the figure to scale on coordi- 
nate pap*er, using a protractor to lay off correctly to degrees the 
angles given, before attempting to apply any formulas ; then 
write the required equations directly from a consideration of 
the figure, and not by attempting to memorize the solutions 
for the different types of problems. The lengths as given 
graphically serve as a check upon the values obtained. 

206 



RIGHT TRIANGLES 



207 



2. Right triangles. Type I. — Given hypotenuse and one 
angle, i.e. a and r, or ft and r. 

A guy wire 168 feet long reaches to the top of a tall chimney, 
making an angle of 37° with the ground. Find the height of 
the chimney and the distance 
from the supporting peg to 
the foot of the chimney. 

Solution. — First draw the fig- 
ure, as indicated, using \ inch to 
represent 30 units. 

y = 168 sin 37° 
x = 168 cos 37° 
Check, y = x tan 37° Hypotenuse and one angle given 

Using natural functions there are here three problems in multiplication. 
The logarithmic solution is as follows : 




log 168 = 

; sin 37° = 

log y - 



2.2253 
9.7795- 10 



= 2.0048 
= 101.1. 



log 168 = 

log cos 37° = 

logx = 

X = 



2.2253 
9.9023 



10 



2.1276 
134.2. 





H 

* 


* 

-4— 


B- 



Angle and side given 



Check, log x = 2.1276 

log tan 37° = 9.8771 — 10 
log y = 2.0047. 

Compare with preceding value of log y ; a dis- 
agreement in the fourth place is permissible. It 
would not affect the fourth significant figure of 
y in this case ; nor would the measurements of 
height of chimney and length of guy wire be 
made with greater accuracy than to one tenth of 
a foot, x 2 + y 2 = 168 2 could be used as a check. 

3. Right triangles. Type II. — Given 
one leg and an angle. 

If a telegraph pole is 34 feet high, and 
the supporting wire makes an angle of 62° 
with the ground, find the length of the 
wire and the distance from the foot of the 
supporting peg to the foot of the pole. 



208 



UNIFIED MATHEMATICS 



The corresponding formulas are as follows : 



r = 



34 cot 62 ( 
34 



Check, x = r cos 62° 
or 34 2 = r 2 - x 2 

= (r — x)(r + x) 
log 34 = 11.5315-10 
- log sin 62° = 9.9459 - 10 
los: r 



sin 62° 
log 34= 1.5315 
log cot 62° = 9.7257 - 10 

logx = 1.2572 logr= 1.5856 

x = 18.08. r = 38.51, 

Check, logr = 1.5856 
+ log cos 62° = 9.6716- 10 

logo? = 1.2572, checks. 



4. Right triangles. Type III. — Given a leg and the hy- 
potenuse. 

Let the side a = 341 and c, the hypotenuse, equal 725. 
Evidently, sin a = f f 1, . 

b = 725 cos a. 



log 341 

log 725 

log sin a 



12.5328 

2.8603 



10 



9.6725 
a = 28° 4'. 



10 



:4Q 



40 



=E?$ 



ff fftfftttffl 



:igO30 



ck. b = 341 cot a. 




log 725= 2.8603 




log cos a= 9.9457- 


-10 


log b = 2.8060 




b = 639.7. 




C%ec&. log 341 


= 2.5328 


log cot a 


= .2731 


log & 


= 2.8059. 



Hypotenuse and one side given 

Graphical solution gives a rough check 
to two significant figures. 



Compare with above value 
log b =2.8060 as a check; 
the error of 1 here is in- 
evitable with 4-place log- 
arithms. 

Note that on the small 
graph only two places are 
accurately representable. 



5. Right triangles. Type IV. — Given the two legs. 

Entirely similar except that the initial formula is for tan a instead of 
sin a. 

Note that commonly in lettering right triangles x and y or a and b are 



RIGHT TRIANGLES 209 

used for the legs, r or c for the hypotenuse, a and /3 for the angles at A and 

5, opposite a and b respectively. 

In Type III if a and c are nearly equal we may avoid the use of the 
sine of the angle near to 90° hy computing the other side using the 
formula 

b°~ =c 2 -a2 = (c- a)(c+ a). 
Thus if a = 718, c = 725, b* = (725 - 718) (725 + 718) 

= 7(1443) = 10101. 
b = 100.5, 
either by inspection as in this case, or from a table of squares, or by 
logarithms. 

PROBLEMS 

Solve the following right triangles by logarithms : 

1. Given r = 240, a = 30° 10'. 

2. Given a = 368, a = 30° 14'. 

3. Given a = 368, r = 579. 

4. Given a = 368, b = 275. 

5. Solve for the missing parts the following ten problems, 
using logarithms ; time yourself ; the exercise should be com- 
pleted within 30 minutes. 

a. Given r = 186, a = 84.3. e. Given a = 930, a = 24°. 

6. Given a = .394, b = .654. / Given b = 184, a =55° 15'. 

c. Given a = 2.89, /? = 68° 24'. #. Given r = .0936, 6 = .0418. 

d. Given 6 = 706, a = 70° 10'. h. Given 6 = 3.24, fi = 86° 14'. 

*. Given b = 878, a = 48° 19'. 

j. Given r = 8.4 x 10 6 , /? = 34° 16 r . 

6. Area. — In computations of functions involving measured 
and computed values, measured values are taken, as far as 
possible, in preference to computed values. The computed 
value involves not only the inaccuracies or errors of measure- 
ment, but also the errors of computation, the inevitable errors 
of computation with approximate numbers as well as the 
avoidable errors. Among the following formulas for the area 
of a right triangle the student should select, in accordance with 



210 



UNIFIED MATHEMATICS 



the principle mentioned, the formula to be used in each prob- 
lem involving a right triangle. 



A = \ ab = \ a 2 cot a 



4- b 2 tan a = \ 



c 2 sm a cos a. 



7. Applications. — In the application of the solution of 
right triangles to practical problems we find that the difficulty 

is frequently a matter of ter- 



H- 



m 



1,1 I I I u 



Ba 



Common terms relating to angles 

Dip — depression — elevation — bias 
— departure. 



minology rather than of prin- 
ciple. The student is urged 
to acquire some real famili- 
arity with the industrial and 
scientific application of the 
principles explained. 

The terms " elevation," 
" depression," " dip," " de- 
parture," and " bearing," all refer to angles. Thus in the figure 
ABC, if AC is in the direction of the sun or if C represents the 
top of a mountain viewed from A, then angle B AC is termed the 
angle of " elevation " of C as viewed from A ; if the observer 
is at C, on a mountain or in an airship, HCA is the angle of 
"depression" ; if A represents the horizon as ^viewed from (7, 
then HCA is called the " dip " of the horizon. If CA repre- 
sents a vertical section of a vein of coal, the angle HCA or 
BAC is called the " dip " of the vein ; in navigation if AB 
represents east, then angle BAC represents the " departure " 
north of the line AC, whereas 
in surveying the angular de- 
flection from north or south is 
given as the " bearing." 

Frequently some function 
of an angle is given from 
which the angle must be de- 
termined. The pitch of a roof 
is given as the height divided 
by the span, whence the corresponding slope angle of the roof 
is 0, given by 



-p 


^X - 


■**■ - 4^ 


:: ^___±_s s 


M 1 / nil ffr 


:::::-?-: :::: Z ::::::^"::::: 


--•<§: :::: ::::::::*S"= 


::^-:l: - 3s~ 


::I::::z-::::S:::::::::::5i:: 


::::+::::::::::::f:: 


--< -^s — .^ho/n \>— 


4-rHfm -F — 



Pitch 



h_ 
2s 



RIGHT TRIANGLES 



211 



tan 6 = -, 

s 

where h is the height and 2 s is the total span. 

The slope of a railroad is commonly given as so many (h) 
feet of rise in 100 feet horizontally ; this gives the slope angle 

from tan = -^-=* %. 

A spiral thread winds about a cylinder advancing a height 
h, called the " lead," in one complete turn ; the circumference 




Full size representation of a one-inch cylindrical screw 

The " lead " is 2 8 o oi an inch. 



of the cylinder is the base and the " lead " is the altitude of 
a right triangle which may be regarded as wrapt about the 
cylinder to give the spiral. The angle a made with h by the 
spiral line is called the angle of the spiral ; evidently 
2_7rr 
h 



tan « = 




Circumference, AC, and length of one spiral, AB, of the above one-inch 
cylindrical screw 



212 



UNIFIED MATHEMATICS 



w-- 



la 



PROBLEMS 

1. A standpipe subtends an angle of 4° at the eye of an 
observer ; if its height is 280 feet above the level of the eye, 
find its distance from the observer. 

2. If the diameter of the standpipe in 1 subtends an angle 
of 15' at the eye, what is the diameter in feet ? Suppose that 
the angle at the eye lies between 10' and 20', what range of 
diameter would these values give ? 

3. The shadow of a flagstaff 60 feet high is 48 feet long. 
Find the angular elevation of the sun. 

4. Using trigonometric functions, find the height of a 
building which at the same time casts a shadow 87 feet long. 

5. Find the lengths of the circle of latitude and the circle 

of longitude through your home 
city. 

6. When the sun is directly 
over the equator, the latitude of 
any place on the earth's surface 
from which the sun is visible is 
the angle betweeli the zenith line 
(the vertical) and the line to the 
sun, when the sun is on the me- 
ridian. Find the length in feet 
of the shadow of a pole 100 feet 
high, at noon, latitude 40° N., and 
also for latitude 42° 18' K 

7. Compute the diameter of a circle circumscribed about 
an equilateral triangle of side 40. 

8. Find in a circle of radius 486 cm. the chords of angles 
of 30°, 60°, 45°, 90°, 120°, 72°, 68°. 

9. For any angle a, find the chord and the chord of half 
the angle in terms of the radius r. Apply the latter formula 
to obtain the results of problem 8. 



Zenith distance representing 
latitude 

Sun on celestial equator, di- 
rection OE. OP is zenith 
direction of P. 



RIGHT TRIANGLES 213 

10. A pendulum of length 34 inches swings between two 
points 10 inches apart ; compute the arc of the swing. If this 
is a seconds pendulum, passing the vertical once every sec- 
ond, what is the velocity of the pendulum bob? Find the 
chord of this arc, and the difference between the chord and the 
arc. 

11. A circular arch over a doorway is to be 4 feet wide and 
20 inches high ; compute the radius. Compute for heights 
of 10 to 24 inches by 2-inch intervals. 

12. Given the radius 10 feet and the span 4 feet of a 
circular arch. Compute the height. Compute for spans of 
2 feet to 20 feet, by 2-foot intervals. 

13. Adapt the preceding results to a radius of 8 feet. How 
closely would interpolation give correct results ? discuss by 
considering the problem graphically, 

14. In a circle of radius 100 inches, compute to one decimal 
place the lengths of sides and the perimeters of regular in- 
scribed polygons of 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12 sides. Time 
yourself on the exercise* State the general formulas involved. 

15. Compute perimeters of regular circumscribed polygons 
of 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12 sides in a circle of radius 
100. Time yourself on the numerical work ; 30 minutes is 
ample time. 

16. Compute the perimeter of a regular inscribed polygon 
of 96 sides, and of a regular circumscribed polygon of 96 
sides, radius 100. How does the circumference compare with 
these two values ? Archimedes computed these lengths by 
plane geometry methods and so found ir to lie between 3^- and 
314/ Check his result. 

17. Frequently arches of bridges are circular segments; 
find the radius of the circular arch of the famous Rialto in 
Venice (see illustration, page 225). The width of the arch is 
95 feet and the height is 25 feet. Draw the graph of the arch 
to scale. 



214 



UNIFIED MATHEMATICS 



18. One of the largest masonry bridges in the U. S. is the 
Rocky River bridge at Cleveland ; the height of the circular 
arch is 80 feet and the span is 280 feet. Find the correspond- 
ing radius. 

19. Find the angle of the spiral represented in the above 
diagram of the one-inch cylindrical screw. 

20. Find the angle of a cylindrical screw of diameter i inch 
which advances -fa of an inch in one complete turn. 

21. A twelve-inch gun has a muzzle velocity of approxi- 
mately 2500 feet per second (f.s.). The velocity is tested by 




Determination of velocity ov a projectile 
One screen is at the muzzl of the gun. 

electrical means ; screens are placed at a known distance apart 
and the projectile in passing through the screens breaks 
successively two electrical circuits which serve to give the 
time of flight of the projectile to thousandths of a second 
in passing through the known distance. The apparatus may 
also be used to determine the angle of elevation of the larger 
guns. In the figure TAT' represents the axis of trunnions of 
a twelve-inch gun ; AM along the axis of the barrel is 25 feet ; 
MS is 180 feet ; the one screen is over the muzzle and the 
other screen is at a height of 94 feet above the axis of trun- 
nions. Determine the angle of elevation of the gun and 
reduce to " mils." Find the horizontal distance MQ between 



RIGHT TRIANGLES 



215 



muzzle and screen and the vertical distance between muzzle 
and screen, QS ; find the time of flight of the projectile, 
assuming 2500 f.s. as velocity; And "horizontal velocity," v., 
and " vertical velocity,'' v y , by dividing MQ and QS re- 
spectively by this time of flight. The vertical velocity v y 
divided by 32.2 gives approximately the time in seconds that 
the projectile will continue to rise ; find this time ; the position 
of the projectile after this interval of time is given approxi- 
mately by the product of horizontal velocity, v x , multiplied by 
the time, as horizontal distance from the gun, and by vertical 
component of velocity multiplied by the same value of t less 
16.1 multiplied by t 2 , as ordinate. The equations are 
x = vj, 

y = Vy t - lQ.lt 2 . 

What error is possible in the angle measured if the height of S 
is given only within one foot ? The aim is directed at a point 
two feet below the top of the screen, as, in general, there is a 
slight " jump " due apparently to the explosion. Estimate 
the jump in degrees and minutes, and in " mils " if the pro- 
jectile hits the top of the screen. 

22. When two screens are used with a large gun the dis- 
tance between screens is sometimes measured by taking equal 




Distances of screens from muzzle, M, determined by right triangles 
Two screens spaced 100 feet apart (horizontally). 

distances MH and ME at right angles to the line MS ± S 2 and 
measuring the angles MH8 U MHS 2 , -MES l9 and MES 2 . Note 
that the screens are 20 to 100 feet in the air on tall standards, 



216 



UNIFIED MATHEMATICS 



making it inconvenient to measure the distance with a steel 
tape. Assuming that the distances ME and MH are taken as 
20 feet and that the angles MHS X = MES l = 70° 10', and that 
angle MHS 2 = MES 2 = 82° 34', compute the distance MS l9 
MS 2 , and S l S 2 . If the screen S^ is at an elevation of 34 feet 
and the screen S 2 is at an elevation of 83 feet, compute the 
angle of elevation of the gun, and the height of the muzzle 
above the plane of its axis if the muzzle is 25 feet long from 
the axis. 

Note that the relative positions of the screens are usually 
determined by two observers in towers whose distance apart 
is fixed ; these observers record positions of muzzle and each 
of the screens. 

23. If a stick of length 12 units casts shadows of lengths 4, 
6, 8, 10, 12, 15, 18, 30, and 40 units respectively, determine the 
angle of inclination of the sun. For angles of inclination of 
10° to 20° by degrees, determine the corresponding shadow 
length to tenths of one unit. This type of table was the first 
appearance of the cotangent function as direct shadow; it 
appeared as early as 900 a.d. in the works of the great Arabic 

astronomer Al-Battani. 

24. The pitch of a roof is given 
by the vertical height h, from 
the point C to 31, on the diagram, 
divided by the span, 2 s ; thus -J- 
pitch is a 45° slope. Find the 
slope angle of a roof of -| pitch, 

of I pitch. If 2 s is given as 48 feet, find the length of the 

rafters in each of the roofs mentioned. 

25. In a roof of span 62 feet find to the tenth of an inch 
the lengths of the rafters if the roof is inclined at 30°, 40°, 42°, 
45 9 , 53°, and 60°. In each case determine the effect upon the 
length of the rafter of an error of one degree. 

26. Find the pitch and the angle of inclination of the roof 
represented in the diagram above. 



E|:::::EEEEE|::::::^:^^ 




-± „-^_- S 4 £-'SS;-"- 


MglffilSri 



Pitch equals — when span is 2 s 



RIGHT TRIANGLES 



217 



8. Railroad curves. — In so far as possible the track of a rail- 
road is laid out in straight lines. Wherever the direction of 
the track is changed a curved line of track is introduced lead- 




Simple curve at a turnout on a railroad track 



ing from the one straight track to a second; these straight 
portions of track must be tangent to the curve which joins 
them, and so they are commonly designated simply as tangents. 
Let AV and VB in the figure represent two such tangents, 
meeting at a point V, called the vertex ; the exterior angle 
XVB is called the deflection 
angle, and is usually designated 
by I. A single circular arc, 
radius R, which joins two tan- 
gents is called a simple curve and 
is designated in American rail- 
road practice by the number of de- 
grees D at the center of the circle 

subtended by a chord whose length is 100 feet, the length of 
one chain used in surveying. On a simple curve the lengths 



_____ 




■_t 




± 








- - _t _?!._''■ 




_t=6»" 


— " ^ " H ^ 


,.--:|f 1} _- 






1^-" 






"5 ^ttH 




A 


B 


li II 


_u 



Chord 100 feet; arc approxi- 
mately 100 feet 



218 UNIFIED MATHEMATICS 

of two consecutive tangents, from intersection point to the 
circle, i.e. AV and VB, are equal ; this length is called T. 
The angle D is commonly given only in degrees and half- 
degrees. 

Relation between D and R. Let PB on the figure represent 
a chord of length 100 feet ; drop the perpendicular from A, the 
center of the circular arc of radius R, bisecting PB. Evi- 

D 50 50 

dently sin — = — , whence R = -. Now for any angle 

2 R • D 

sin— 

2 
up to 4° the sine differs numerically from the angle expressed 
in radians by less than .1 of 1% of itself; hence you may 

replace sin — by — • -^— , the value of — in radians, with an 

F 2 J 2 180 2 

error of less than .1 of 1 % when D is any angle up to 8°. 
Note that the error is less than 5 feet in 5000 ; the circular 
measure of the angle is larger than the sine so that the error 
will be a deficiency. 

D 18000 

XV = 

ttD 

gives the radius. 

Relation between I, R, and T. On our figure in the right 

triangle OAVthe angle AVO is 90° — -, and the angle AOB is 

evidently equal to the deflection angle I. 

\ 2 J AV T 9 

whence R = T cot - , and T = R tan £. 

Evidently the radius R can be expressed in terms of the 
"degree" D of the curve, giving new formulas involving 
D, T, and i". 

Elevation of outer rail. In turning a curve a railroad train 
tends to leave the track, due to the tendency of any moving 
body to continue its motion in a straight line. To keep the 
train on the track the flanges alone are not sufficient, but the 



RIGHT TRIANGLES 219 

outer edge must be elevated. The formula for ordinary speeds, 

giving number of inches of elevation, is e = -& — , wherein g is 

32 R 

the gage of the track in feet, v the velocity of the train in feet 
per second, and R the radius of curvature in feet. 



PROBLEMS 

1. What radii have railroad curves of 8°, 7°, 6°, 5°, 4°, 3°, 
2°, and 1°, respectively ? 

2. If a railroad curve is built with the radius of 2640 feet, 
compute D in degrees. 

3. On a circular track of 100 miles' circumference what 
would be the number of degrees ? 

4. On English and continental railroads the curvature is 
usually given by the length of the radius ; find the number of 
degrees, American D, corresponding to radii of 8000, 5000, 
4000, 3000, 2000, 1000, 800, 600, and 400 feet, respectively. 
Do not compute beyond minutes. Find D for radii of 300 
meters, 1000 meters. 

5. Compute e, elevation of outer rail, for g = 4 feet 8.5 
inches, standard gage on American railroads, when v = 60 
miles per hour, and R = 800, 1000, 2000, 4000, and 5000 re- 
spectively. Compute for a one-degree and for a two-degree 
curve. 

6. Given that two portions of straight track diverge at 
22° 14', and that the tangent distance is to be 300 feet, com- 
pute R ; find R for T, the tangent distance, equal to 200, 250, 
and 350. Eind the corresponding values of D. How could 
you determine the length of T, approximately 300, so that 
D will come out in degrees and half-degrees ? 

7. Compute R when T = 400, 500, and 600, respectively, 
the deflection angle being 60° ; similarly when I = 30°. 

8. Compute e for g = 4 feet 8.5 inches (4.71 feet), standard 
gage, v = 60 miles per hour, and curves of 1°, 2°, 5°, 6°, and 8°> 
respectively. 



CHAPTER XIV 

THE CIRCLE 
1. Formulas. — 

x 2 +?/ 2 = r 2 . | x — h = r COS 0, Parametric equa- 
(x — h) 2 + (y — k) 2 = r 2 . \ y — k = r sin 0. tions of the circle. 




OM? + JfiPi2 = 0Pi2, OJf 2 2 + lf 2 P 2 2 = OP 2 2, OMf + ^ 3 P 3 2 = OP 3 2, 

OJf 4 2 + J9f 4 P 4 2 = OP 4 2 . 

»i 2 + 2/i 2 = r 2 ; x 2 2 + ?/ 2 2 = r 2 ; z 3 2 + y£ = r* ■ » 4 2 + 2A 2 = r* ; x* + y* = r*. 

For any point P (a?, ?/) on a circle of radius 10, center the 
origin, we have the relation, 

a;2 + ^ = 10 0, 
220 



THE CIRCLE 221 

which is the equation then of a circle of radius 10 and center 
at the origin (0, 0). This equation is obtained directly from 
the distance formula ; x 2 -f- y 2 = 100 expresses the fact that the 
distance of the point (x, y) from the point (0, 0) is 10 ; any 
point (x, y) which satisfies this equation is at a distance 10 
from (0, 0) and any point at a distance of 10 units from O 
■(0, 0) satisfies this equation. The locus of this equation, then, 
is the circle of radius 10 and center (0, 0). 

The formula may readily be verified on the figure ; take 
P any point on the circle, drop PM a perpendicular to the 
a--axis. Then OM 2 + MP 2 = OP 2 , in any one of the four tri- 
angles, representing any possible position of P. Herein OM 
and MP must be regarded initially as positive quantities, since 
the formulas of plane geometry were applied only to positive 
lengths. However OM, as a positive length = x, where the 
negative sign is taken for points in II and III and MP= y, 
where the negative sign is taken in III and IV, whence sub- 
stituting in OM 2 + MP 2 = OP 2 you have x 2 + y 2 = 10 2 . For a 
circle of radius r the equation x 2 -\- y 2 = r 2 is satisfied by any 
point P(x, y) which is upon the circle, for OP will equal r, and 
every point which satisfies the equation evidently lies on the 
circle. Hence, by definition, the locus of x 2 -f y 2 = r 2 is the 
circle of center (0, 0) and of radius r, for every point on 
the circle satisfies this equation and every point which satis- 
fies the equation lies upon the circle. These two conditions 
must be fulfilled in order that any given curve may be desig- 
nated as the locus of a given equation. In other words, the 
given curve must include all points whose coordinates satisfy 
the equation and must exclude all whose coordinates do not 
satisfy the given relation. 

Similarly the two equations : 

x = 10 cos 0, 
y = 10 sin 0, 

give for every value of 6, called a parameter, the coordinates 
of a point which lies upon the circle. The locus of this pair 



222 



UNIFIED MATHEMATICS 



of equations is the circle of radius 10. Thus the ten values of 
= 0°, 30°, 45°, 60°, 90°, 120°,_150°, 180°,_210°, and 330°, give 
the ten points, (10, 0), (5V3, 5), (5V2, 6V2), (5, 5V3), 

(0, 10),_(- 5, 5V3), (- o V3, 5), (- 10, 0), (- 5V3, - 5), and 
(+ 5V3, — 5), which lie upon the circle. Intermediate values 



V&Ml 




Circle of radius 10 ; units are eighths of an inch 

x = 10 cos 0. 



x 2 + y* = 100, or 



y = 10 sin 0. 



can readily be obtained using the tables of sines and cosines. 

The two equations together constitute the equations of the 
circle in parametric form, a type of equation of particular im- 
portance in applied mathematics. 

If desired, we may eliminate 6 as follows : squaring and 
adding gives 

X 2 + y 2 = 10 (cos 2 6 + sin 2 0), 

or x 2 + y 2 = 100 (since sin 2 6 + cos 2 = 1), 



THE CIRCLE 223 

a relation independent of 6. But for many purposes it is more 
convenient to keep the equations in parametric form. 

For the distance from any point C (h, k) to a point P (x, y) we 
have found the formula d = V(# — h) 2 -4- (y — k)' 2 ; all points 
(x, y) which satisfy this equation for a given value of d, and 
for (7i, k) a fixed point, lie upon a circle of which (h, k) is the 
center and d is the radius ; no point not on the circle satisfies 
this equation. 

(x — h 2 ) + (y — A:) 2 = r 2 is then the equation of a circle of 
center (h, k) and radius r. Any equation which can be put 
into this form represents a circle, for it expresses the fact 
that the distance from any point (x, y) whose coordinates 
satisfy the given equation, to the fixed point (Ji, k) is constant 
and equal to r. 

In parametric form, the two equations representing the 
circle with center (Ji, k) and radius r are written : 
x — h = r cos 9. 
y — k = r sin 0. 

If is given values the corresponding values of x and y 
determine points upon the circle (x — h) 2 + (y — k) 2 = r 2 . 

Illustrative problem. — Find the equation of the circle 
of radius 5 ; center (3, — 7). 

By the distance formula, taking (x, y) as any point on the circle, 
(x-3) 2 +0/ + 7)2 = 25, 
or x 2 - 6 x + y 2 + 14 y - 33 = 0. 

In parametric form the equations of this circle are, 
x — 3 = 5 cos 6. 
y + 7 = 5 sin 6. 

2. Reduction to standard form. — Any equation of the type, 
a-2 + y 2 + 2 Gx + 2 Fy + G = 0, 

or Ax 2 + Ay 2 4- 2 Gx + 2 Fy + C = 0, 

represents a circle. The center and radius are determined by 
completing the square, as in the illustrative problem below. 
If the expression for the radius is zero, the circle reduces to a 
point ; if it is negative the circle is imaginary. 



224 UNIFIED MATHEMATICS 

Illustrative problem. — Find the center and radius of the circle, 

2 x 2 + 2 ?/ 2 + 6 x - 7 y - 15 = 0. 

This equation represents a circle since it can be put into the form of a 

circle, as indicated herewith : 

2(z2 + 3 x) +2(y*-|j0 =15. 

2(x 2 + Sx + f) + 2(2/2 _ |y + fl) = 15 + | + -V-. 

2(ai + D* + 2(y - |)» = Jfi. 

(x + I) 2 + (y - |) 2 = -W (or 12-81). 

-v/905 
This equation states that the point (x, y) is at the distance from 

the point ( — §, |) ; this equation represents a circle with the center 

(- |, |) and radius -^^ or l^ 2 or 3.58. 



PROBLEMS 

Find the equations of the following circles : 

1. Center (3, — 4), radius 5. 3. Center (— 4, 0), radius 4. 

2. Center (0, 0), radius 10. 4. Center (—6, 6), radius 6. 

5. Center (—6, — 8), radius 10. 

6. Draw the circle of radius 10, center (0, 0) and estimate 
carefully its area on the coordinate paper. 

Find the centers and radii of the following circles ; time 
yourself ; the eight problems should be completed numerically 
within 12 minutes. 

7. X 2 + y i _12 x + 12 y + 36 = 0. 

8 . X 2 + y i _ 12 x + 12 y 4- 36 = 0. 

9. a; 2 -|-2/ 2 -39 = 0. 

10. a; 2 - 10 a; +?/ 2 - 39 = 0. 

11. 2 a; 2 -f 2 if - 6 a; - 8 y - 19 = 0. 

12. 2 a; 2 -f 2 ?/ 2 - 5 a; + 7 */ - 15 = 0. 

13. 3 x> 4- 3 */ 2 - 15 x 4 17 y 4 9 = 0. 

14. a? 2 4 6 a? 4- y 2 - 10 = 0. 

15. Draw the graphs of the preceding 8 circles, using only 
one or two sheets of graph paper ; time yourself, keeping a 
record of the time. 



THE CIRCLE 



225 



16. Given 



x = 5 cos v, 



y = 5 sin 0, 
locate 16 points on the curve, using the values sin 0° = 0, 
sin 30° = .5, sin 45° = .707, sin (50° = .866 for these and related 
angles. 

17. Given x = 5 + 5 cos 0, 

y=z — 3 + 5 sin 0, 
locate 16 points on this circle. 

18. When 0= 37°, 43°, 62°, 80°, and 85° find x and y in the 
preceding problems. 

19. Through what point on the circle x 2 + y 2 — 25 does the 
radius which makes an angle arc tan 2 with OX, pass ? 




The Rialto in Venice 
A famous circular arch, 95 feet wide by 25 feet high. 

3. To find the intersection of a line with a circle. Tangents. — 

The intersections of the circle, x 2 + y 2 = 100, with any line 
as y = x + 5, are represented by the solutions of the two equa- 
tions regarded as simultaneous. Six problems are given here. 
1. #2 + 2/2=100, 2. » 2 + y 2 =100, 

y = x. y = x + 5. 



226 



UNIFIED MATHEMATICS 



3. x* + y 2 = 100, 
y = x_+ 10. 

4. x* + f- = 100, 
y = x + 16. 



5. a; 2 + ?/ 2 = 100, 
2/ = x — 8. 

6. x 2 + ?/ 2 = 100, 
?/ = a; + fc. 



Solving, by substitution in each of the six cases indicated 
above : 




Graphical solution, determining intersections of the circle, jc 2 + y 2 = 100, 
with various lines of slope 1 

1. gives 2 x 2 = 100, x 2 = 50, x = ± V50 = ± 7.07, y = ± 7.07 ; 

2. gives 2 x 2 + 10 x + 25 = 100, x 2 + 5 x - 37.5 = ; 

x = _ 2.5 ± V6.25 + 37.5 = - 2.5 ± 6.61 

= + 4.11 or -9.11, 
y = 9.11 or -4.11; 

3. gives x 2 + 10 x = 0, x = or — 10 (by factoring, simplest), 

y = 10 or ; 



THE CIRCLE 227 

4 gives 2 x 2 + 32 x+ 156 = 0, x 2 + 16 x 4- 78 = 0, 
x = -8 + V 64 - 78 
= _8+V-14 
= imaginary values, not any scalar values of x; 

5. gives 2 x 2 - 16 x - 36 = 0, x 2 - 8 x - 18 = 0, x = 4 ± V34 

= 4 ± 5.83 
= 9.83, or - 1.83, 
y = 1.83 or - 9.83 ; 

6. gives 2 x 2 + 2 k • x + (fc 2 - 100) = 0, 

= - fc ± Vfc 2 - 2(fc 2 - 100) 

2 



=— -±i V200- P. 

Very evidently the solutions of 1 to 5 are all included under 
the solution 6, as special cases. 

Geometrically the lines of slope 1 are divided by the circle 
into three classes, viz. (a) those which cut the circle in two 
distinct points : (b) those which do not cut the circle ; and (c) 
those which are tangent to the circle, or cut the circle in two 
coincident points. 

Evidently lines in 1, 2, 3, and 5 belong to the first class ; the 
line in 4 to the second class. To determine the tangents one 
must fiud the value of k for which 200 — k 2 = 0, as only when 
200 — k 2 = are two points whose abscissas are given by 

- £ + \ V200 - A- 2 and -\-\ V200 - A? coincident. In this 

case, k = ± 14.14, the lines y = x ± 14.14 are tangent to the 
circle x 2 + y 2 = 100. The abscissa of the point of tangency is 
+ 7.07, since it equals 

Rule. — To find the tangent with given slope to a given circle 
write the equation of the family of lines of the given slope, 
y = mx 4- k, and solve for the points of intersection with the circle; 
get the condition that the two points of intersection should be coin- 
cident. This gives the value of k for which the line y = mx 4- k 
is tangent to the given circle. 

Note. — The method will apply to any curve of the second degree. 



228 



UNIFIED MATHEMATICS 



4. Circles satisfying given conditions. — To find the equation 
. of a circle which satisfies given conditions it is necessary to use 
the analytic formulas which we have derived combined with 
the geometric properties of a circle. In general call (h, k) or 
(aj, y) the center of the circle and r the radius ; sketch the lines 
and points which are given and indicate roughly the probable 
position of the desired circle ; solve the problem geometrically 
if possible, or indicate the solution, and express the geometrical 
facts in algebraical language by using the preceding formulas. 



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EEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEffiES 



Circle through three points 

Determination of center by perpendicular bi- 
sectors of chords. 



5. Illustrative 
problems. — Find 
the equation of the 
circle through A 
(1, 2), B (0, 8), and 
G (7, - 1), (1) using 
the distance formula, 
(2) using the per- 
pendicular bisector 
of the line joining 
two points, (3) using 
the general equation 
(x-h) 2 + (y-lc) 2 =r* 
which may represent 
any circle, and (4) 
using the general 
equation 

Ax 2 + Atf + 2 Gx 
+ 2 Fy + C = 0. 



(1) Call the center P(h, k), then PA=PB, PB = PC, and P_4=PC. 

The distance from A to P equals the distance from # to P, whence by 
the distance formula, 



V(/i - 1)2 + (& - 2)2 =: V(/l - 0)2 + (fc - 8)2 



THE CIRCLE 229 

Similarly, 

II. y/(h - 1) 2 + (k - 2) 2 = v'(A - 7)2 -f- (& + 1)2 

expresses analytically the fact that P4 = PC ; and 

III. v'O - 0)2 + (k - 8) 2 = V(/t - 7) 2 + (A: + 1)2 
that PB= PC. 

Since equation III is derivable from I and II, it adds nothing new ; 

any two of these equations are sufficient to determine (h, k) the center. 

Squaring in each and combining terms we obtain from I, 

12 k - 2 h — 59 = 0, 

a straight line which is the locus of all points equidistant from A and B ; 

and from II, . 7 , ., - n 

4 A — 2 1 & — 15 = 0. 

Solving, we obtain the one point which is equidistant from A, B, and C, 

h = 6.77. 

k = 6.05. 



r = V(6.77 - l) 2 + (6.05 - 2)2 = v'(33.29 + 16.40) = V4^69 
= 7.05. 
The circle is (x - 6.77) 2 + (y - 6.05)2 _ (7.05) 2 . 

(2) The center of the circle is the intersection of the perpen- 
dicular bisectors of the sides ; finding the slopes of the sides, 
the rnid-points, the slope of the perpendicular to each side, the 
equations of the perpendicular bisectors of AB and AC are 
found (point-slope) to be 

12 y - 2 x - 59 = 0. 
Ax-2y -lo =0. 

Examination shows that these are precisely in x and y the equations 
obtained in our first solution in h and k and from this point the solution 
proceeds as in (1). The student should explain the reason for this. 

(3) I. (x - hf + - £) 2 = r« 
is the equation of any circle, center (h, k), radius r. 

Substituting in this equation (1, 2), (0, 8), and (7, — 1), gives, 

ii. (i - hy + (2 - ky = ?-2. 

HI. (0_/i)2+ (8-A:)2 = r 2. 

IV. (7 - h) z + (- 1- ky - r\ 

V. Ill - II - 2 h + 12 k - 59 = 0. 

VI. II - IV -48 + 12fc + 3-6& = 

or 4 h - 2 k - 15 = 0. 



230 UNIFIED MATHEMATICS 

V and VI are seen to be in h and k precisely the equations solved in 
method (1), and in method (2) for x and y as variables. 

(4) I. Ax 2 + Ay* + 2 Gx + 2 Fy + C = 0. 

Substitute in this equation (1, 2), (0, 8), and (7, — 1) and solve for 
the values of G, F, and C in terms of A. 



II. 




A + 4A + 2G + 4F+ C = 0. 


III. 




64 A + 16 F + C = 0. 


IV. 




49A + A + UG-2F+C = 0. 


V. 


II - III 


-59A + 2G-12F=0. 


VI. 


II - IV 


- 45 A + 12 G + 6 .F = 
or - 15 J. - 4 G + 2 F = 0. 



G F 

These are the same equations in and , regarded as the un 

A A 

knowns, as appeared above in h and k. 

6. Tangency conditions. — If a circle is to be tangent to a 
given line the distance formula (normal form) from a point to 
a line may be used ; if a circle to be found is to be tangent to 
a given circle, then the radius sought, plus or minus the given 
radius, must be equal numerically to the distance from center 
to center, according as the circles are tangent externally or 
internally. 

7. Circle through the intersection of two circles. — 

I. x 2 + y 2 + 10 # = 0, a circle of radius 5, center (—5, 0). 

II. x 1 + y 2 — 49 = 0, a circle of radius 7, center (0, 0). 

III. (x 2 + y 2 + 10 x) + k(x 2 + y 2 - 49) = 0. 

The third equation is satisfied by the points of intersection 
of curves I and II, for all values of k (see page 83). For all 
constant values of k, III may be written 

(1 + k)x 2 + (1 + k)y 2 + 10aj - 49&= 0, 

and the form shows that this represents a circle. To deter- 
mine the circle through the intersections of I and II, and any 
other given point substitute the coordinates in III, and solve 



THE CIRCLE 



231 



for k ; since a circle is determined by the three points, it is 
easily seen that every circle through the two points of inter- 
section of the given circle is included in the family of circles, 
(1 + fc) x 2 + (1 + k) y 2 + 10 x -4S)k = 0. The method of deter- 
mining k to have the circle pass through some other given 




Common chord of two circles or radical axis 

point is precisely the same as in the similar problem with 
straight lines (page 83). 

For k = — 1, this equation reduces to the linear equation 
representing the common chord of the family of circles ; 
whether two given circles intersect or not, this line, whose 
equation is obtained by eliminating x 2 + y 2 between the two 
given equations, is called the radical axis of the two circles. 



8. Geometrical property of the radical axis. — 

(a? — h) 2 + (y — k) 2 is the square of the distance from (x, y) 
to the center of any circle ; {x — h) 2 + (y — k)- — r 2 is the 
square of the length of the tangent to the circle from any 
point outside the circle of center (h, k), radius r. 

r <& + V 1 + % Ox + 2 Fy + C is the square of the length of the 



232 



UNIFIED MATHEMATICS 



tangent from any point (x, y) to the circle whose equation is 
x 2 + y 2 + 2 Gx + 2 Fy + C= 0, since the left-hand member is 
identical with the left-hand member when written in this 
form : 

0+ Gf + + Ff -(G* + F*- C)=0. 

Note that if any secant PAB is drawn through P(x, y) then 
P^. • PB = PT 2 ', hence the expression 

x 2 + y* + 2Gx + 2Fy + C 

gives the product of the two distances along any straight line 
from the point P(x, y) on the line to the circle. There is a 




Distances from a point to a circle 

On any secant through P, PAB, PA x PB is constant. 

PA x PB = P2\ 2 = (x - hy + (y - A;) 2 - r*. 

correspondence to the normal form of a straight line, since the 
left-hand member there also represents a distance. 

a* + 2/ 2 + 2 6?^ + 2F 1 y + C l = a? + ?/ 2 + 2 6r> + 2 JF# + C 2 
is an equation which is satisfied by any point from which tan- 
gents drawn to the two circles 

X 2 + yt + 2 G x x + 2 F x y + Q x = 0, 
X 2 + y2 + 2 ^ + 2i^/ + C 2 = 0, 



THE CIRCLE 



233 



are equal in length. Hence, the radical axis is the locus of 
points from which the tangents drawn to the two circles are 
equal in length. 

9. Radical center of three circles. — Given three circles, each 
of the three pairs of circles which may be formed from the 




Radical axes and radical centers 
Radical center of the three circles, 1,2, and 3. 
Radical center of the three circles, 1, 2, and 3'. 



three has a radical axis ; the three radical axes pass through a 
common point, as may be easily shown by Sec. 4, Chapter V. 

a. x* + y 2 +2G 1 x + 2F i y+C 1 = 0. 

b. z 2 + y 1 + 2 G 2 x + 2F 2 y + C 2 = 0. 

c. * + f + 2G0 + 2Fa+Q=O. 

d. (a - c) 2(# x - G 2 )x + 2(F X - F 2 )y+ Q x - C 2 = 

radical axis of a and b. 

e. (6-c) 2(G 2 -G 3 )x + 2(F 2 -Fs)y+C 2 -C 3 =0 

radical axis of b and c. 



234 UNIFIED MATHEMATICS 

/ (d + e) or (a - c) 2(Q 1 ^GJx+2(F 1 -F t )y+C 1 -*C t =Q. 

Since d + e = gives a straight line through the intersec- 
tion of d and e, and since d + e = gives the radical axis of a 
and c, the latter line passes through the intersection of the 
two former radical axes. 

10. Limiting forms of the circle equation. — 

(x — h) 2 +(?/ — k) 2 = r 2 represents a real circle when r 2 is 
positive. 

(x — h) 2 -\-(y — k) 2 =. represents a point circle ; the only 
real point which satisfies this equation is the point (h, k). 

(x — h) 2 -\-(y — k) 2 = — r 2 , r a real quantity, represents an 
imaginary circle ; no real point satisfies this equation, since 
every real value of x and y makes (x — h) 2 positive and (y—k) 2 
positive. 

PROBLEMS ON THE CIRCLE 

1. Find the center and give radius to 1 decimal place of 
each of the following circles ; plot ; find the three radical axes 
and the radical center. 

a. x 2 + y 2 +'6x-8y-16 = 0. , ■ 

b. 3x* + 3y 2 -$x + 15y-7 = 0. 

Hint. 3(x 2 — f x ) + 3(y 2 + 5 y ) = 7 ; complete squares inside 
parentheses and note that 3 times the quantity added within each of the 
parentheses must be added on the right. 

c. ^_j_ 2/ 2_6a;-8 = 0. 

2. Plot the following two circles and determine their 
common chord ; what is its length ? 

a. x 2 + y 2 - 10 x- 100 = 0. 
&.. x' + y 2 + 10y -100 = 0. 

3. Write the equation of the family of circles 

a. With center on #-axis, passing through the origin. 

b. With center on y-axis, passing through the origin. 

c. Passing through the origin. 



THE CIRCLE 235 

d. With center on 3 x — 4 y — 5 = 0, radius 5. 
Note. Sh — 4 ft - 5 = 0. 

4. What limitation is imposed upon the coefficients A, G, 
F, and C in ^ 2 + Ay 2 + 2 Gx + 2Fy+C=0, 

a. if the circle passes through (0, 0) ? (1, 1) ? 

b. if the circle has its center on the axis of x ? ?/-axis ? . 

c. if the circle is tangent to the #-axis ? ?/-axis ? tangent to 
fc-3 = 0? 

(i if the circle is tangent tox — y — 5 = 0? 

5. Find the equations of the circles through the following 
three points : 

a. (0, 0), (6, 0), (0, 8). 

6. (1,5), (-3,1), (7, -3). 

c. (0, 0), .(8, 2), (15, — 3) ; use two different methods. 

6. Find circle tangent to 3 x -\- 4 y — 25 = 0, and passing 
through (2, 3) and (5, 1). Note the two solutions. 

7. Find the radical axis of each of the three pairs of 

CilCleS X 2 + y2_ 6x _ S y-10=0, 

x 2 + y*-20x + 50 = 0, 
2 x 2 + 2 if 4- 8 x + 6 y - 25 = 0. 
Find the radical center. Plot. 

8. Find the tangents of slope 2 to the first circle in 7 ; find 
the normal and the point of tangency. 

9. Find the circle of radius 5 tangent to the line whose 
equation is 4 x — 3 y — 9 = at (3, 1). 

10. Find for what value of r the line 4# — Sy — 9 = is 
tangent to x 2 + y 2 — r 2 = 0. Two methods. Find the point of 
tangency. 

11. Find to one decimal place the points of intersection of 
the circle x 2 + y 2 — 20 x + 50 = with the line y = 2 x — 12. 
Plot. 



236 UNIFIED MATHEMATICS 

12. Use the trigonometric functions to find points of inter- 
section of 

Note that tan d = 2, where 6 is the slope-angle of the line. 

13. Use trigonometric functions to find k, when y = 2 x -\- k 
is tangent to the circle x 2 + y 2 = 100. Draw figure ; note that 
tan = — \ where 6 is the slope-angle of the normal. 

™ i -^ i f a = 3 + 10 cos 0, 

14. Plot the circle < . K , . A . 

{ 2/ = — 5 -f 10 sin 0. 

-.,,,, • , [ a; = 8 sin 0, 

15. Plot the circle I Q 1 

\y = 8 cos 6. 

Note that tf is here the angle made with the y-axis by any radius. 

16. Find the equation of the complete circle of the circular 
arch of the Bialto, referred to the horizontal water line and 
the axis of symmetry of the arc as axes. The arch is 95 feet 
wide by 25 feet high. 

17. Find the equation of the circle of which the arch of the 
Eocky River Bridge, 280 feet by 80 feet, is an arc, referred to 
a tangent at the highest point of the arc as a>axis and the 
perpendicular at the point of tangency as y-axis. Determine 
the lengths of vertical chords between the arc and the .x-axis, 
spaced at intervals of forty feet. 

18. Prove that the radical axis of any two circles is perpen- 
dicular to their line of centers. 

19. If a circle is tangent to a given line and to a given circle, 
what conditions must the coordinates (h, k) of its center 
satisfy ? 



CHAPTER XV 



ADDITION FORMULAS 



1. Functions of the sum and difference of two angles. — The 

formulas for (a + b) 2 and (a — b) 2 are illustrations of addition 
formulas frequently of fundamental importance in mathe- 
matical work. Thus 10* • lO == 10 x+2/ is an addition formula 
leading to the whole theory of logarithms, which revolutionized 
computation processes. The question arises as to addition 
formulas in the case of the trigonometric functions after the 
functions have been defined. Just as the exponent formula 
10 x+y = 10 s • 10 y , which was first proved for positive integers, 
is extended to hold for all values of x and y, so the formulas 
which are established 
for sin (a + P) and 
cos (« + p) when a 
and p are acute 
angles will be found 
to hold for all real 
values of a and p. 



2. Geometrical der- 
ivation of sin (a+P) 
and cos (a + p) ; a 
and p acute and 
a + P<90°. — Given 
a and {$, two acute 
angles whose sum is 
less than 90°, to find 
sin (a + fi) and cos (a + p) 
cos p. 





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237 



238 



UNIFIED MATHEMATICS 



On the figure let a and ft be two positive acute angles whose 
sum is less than 90°, taken, for convenience, distinctly differ- 
ent from each other. Let OP make the angle a with OX, and 
OB make the angle ft with OP, and thus a + ft with OX. 
From B drop perpendiculars B A to OP and BN to OX ; from 
A on OP drop a perpendicular AM to OX ; from A draw a 
parallel to OX cutting BN at R. 

On the figure, noting that OB is taken as r, we have the 
following evident relations : 

AB = r sin ft ; i?5 = AB cos a 
= r cos a sin /? ; 

OA = r cos ft ; Aflf = OJ. sin a 

= r sin a cos ft ; 

.fl"Jf = RA = u4JB sin a 

= r sin a sin ft ; 

OM= OA cos «=?* cos a cos /3. 

. T . s NB MA + #3 

sin (« + /?) = = — • 

r r 

_ r sin a cos ,3 ~h r cos « sin /? 




sin (a-\- ft) = sin a cos ft + cos a sin /?. 

Similarly, cos (« + Q = M= 0^-^ 
r r 

_ r cos a cos ft — r si n a sin /? 



whence 



cos (a + ft) = cos a cos /? — sin a sin /3. 



Having established these formulas geometrically for a + ft 
when < a < 90°, < ft < 90°, and « + < 90°, it now 
remains to establish that these formulas hold for all angles 
a and ft, including negative angles. This extension is made 
by employing the theorems of Section 12, Chapter VII. 



ADDITION FORMULAS 239 

3. Generalization for any two acute angles. — 

sin (« + /?) = sin a cos /? 4 cos a sin /?, 
cos (a + /3) = cos a cos /3 — sin a sin /3. 

First we will show that when a and /? are any two acute 
angles the two formulas established above when a 4- ft < 90° 
continue to hold. The extension to any acute angles requires 
that we prove these formulas to be true further (a) when 
a + (3 = 90°, and (6) when a + < 90°. 

Proo/. (a) If a + = 90°, /3 = 90° - a, whence 

sin /? — sin (90° — a) = cos « ; cos /? = cos (90 — a) = sin a. 

The two formulas then give, by substitution, 

sin (« + /?)= sin 90° = sin 2 a + cos 2 a = 1, 

cos (a + /8) = cos 90° = cos « sin a — sin a cos a = 0. 

The sine of 90° is 1, and the cosine of 90° is ; hence our 
formulas continue to hold even when a -f /3 = 90°. 

(b) a + /8 > 90°. Take the complements of a and /? to be 
respectively x and y, whence x = 90° — a and ?/ = 90° — /?. 
Evidently a; -f- y will be less than 90°, by the same amount that 
a -f (3 exceeds 90°. Further, since x = 90° - a and y = 90° — 0, 

sin x = cos a, sin y = cos /?, 

cos x = sin a, and cos y = sin /?. 

Now, sin (« + /?)= sin (90° - a_+ 90° - ?/) 

= sin (180° — # -f- i/)= sin (a; + y), since 
sin (180° -0) = sin 0. 

Since # 4- ?/ < 90°, sin (x 4 y)= sin a; cos ?/ 4- cos x sin y, as 
established above ; making the substitutions for sin x, cos y, 
cos x, and sin y, we have sin (a 4- /?) = cos « sin /? 4- sin a cos /3. 

Q.E. D. 

Similarly, 

cos (a 4- /?) = cos (180° — a -f y) = — cos (oj 4- 2/), since 
cos (180° — 6)= — cos 0, for any angle 6. But x and */ are 
acute angles, whose sum is less than 90° ; 



240 UNIFIED MATHEMATICS 

therefore cos (x -f- y) — -\- cos x cos y — sin x sin y 

= + sin a sin /? — cos a cos /?. 

Now cos (a -h /?) = — cos (x + y), 
or cos (a + ft) = cos a cos ft — sin « sin ft. q. e. d. 

4. Extension of the formulas for sin (a + p) and cos (a + p) to 
all angles without restriction. — To show that these formulas 
hold for all angles it is necessary now to show that if either a 
or ft is increased by 90° the formulas continue to hold pro- 
vided that they hold for a and ft. 
Thus given 

sin (a + ft) = sin a cos ft + cos a sin ft, 
cos (a + ft) = cos a cos ft — sin a sin /?, 

we wish to show that sin (a + y) and cos (a -}- ?/) are given by 
similar formulas, when y = ft -f- 90°. 

sin (« + ?/)= sin (« + /? + 90°) = cos (a + 0), 
= cos a cos ft — sin a sin ft, 

but sin ?/ == cos ft and cos y = — sin /?, whence substituting, 

sin (a-\- y)= cos « sin y -f- sin a cos ?/. q. e. d. 

Similarly for cos (« -j- y), we find cos a cos y — sin « sin y ; 
since a and ft enter symmetrically in the above formulas this 
proof establishes that a also could be increased or, by an 
entirely analogous procedure, decreased by 90°, with the same 
formulas for the new values. 

This establishes the formulas for any two angles a and ft 
whatever. For since the formulas have been proved above to 
hold for any two acute angles a and ft, the formulas hold for 
any obtuse angle and any acute angle since y, the obtuse angle, 
may be regarded as 90° -f ft. This establishes the formulas for 
any angle in I and any angle in II ; now increase a by 90°, 
thus establishing the formula for any two obtuse angles. 
Continuing in this way a can be any angle in any quadrant, I 
to IV, and ft also an angle in any quadrant whatever, and the 
formulas continue to be true. 



ADDITION FORMULAS 



241 



After these formulas are established for all positive angles 
up to 360°, another method of procedure to establish the 
formulas for all positive and negative angles is to note that 
any integral multiple of 360°, k • 360° with k a positive or 
negative integer, can be added to any angle without changing 
the functions of the angle involved in our formulas. Thus if 
— ft is any negative angle, numerically less than 360°, the 
functions of a-f-(— ft) are the same as the functions of 
a + (360° — ft) which is the sum of two positive angles ; but 
the functions of 360° — ft are the same as those of — ft and 
after application of the formula the 360° can be dropped. In 
other words in these formulas any integral multiple of 360° 
can be added at pleasure and also dropped at pleasure, and in 
this way the formulas are established for all angles. 

Illustrative problem. — Given sin a = .45, cos ft = .68, find 
sin (a + ft) and cos (a + ft). 

sin a = .45 ; a can be in I or II since sin (180° — a) = sin a. 
cos a = ± Vl-.45 2 = ± V.7975 = ± .893. 
cos ft = .68 ; ft can be in I or IV since cos (— ft) = cos ft. 
sin ft = ± Vl - .68 2 = V(.32)(1.68) = ± .16 x V21 
= ±.16 x 4.58 = ± .733. 











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sin a = .45 determines either 
ai or a.2 



cos (3 = -68 determines either 
Pi or p 2 



There are strictly four problems, solved as follows : 

a in I, ft in I. a in I, ft in IV. 

sin a = .45. 
cos a = + .893. - 
cos ft = + .68. 



242 UNIFIED MATHEMATICS 

sin p = + .733. sin p = - .733. 

sin (a + /?) sin (a + p) 

= sin a cos p + cos a sin p. = .45 x .68 — .893 x .733. 

sin (a + p) sin (a + p) 

= .45 x .68 + .893 x .733 = .306 - .655 = - .349. 

= .306 + .655 = .961. cos (a + p) 
cos (a + p) — cos a cos p — sin a sin p. 

= .893 x .68 - .45 x .733 cos (a + p) 

= .607 - .330 = .277. = .893 x .68 + .45 x .733 

= .607 + .330 = .937. 

The two columns represent two solutions which have the three 
central values, sin a, cos a, and cos p, in common. 

The student is expected to complete the solution, beginning 
as follows : 

a in II, p in I. a in II, /? in IV. 

sin a = .45. 

cos p = .68. 

cos a = - .893. 
sin p = + .733. sin p = -'.733. 

In general work only one case, indicating which solution is 
given. 

5. Historical note. — The formulas for sm(a + p) and 
eos(a + P) are closely allied to Ptolemy's theorem (c. 150 a.d.) 
that in any inscribed quadrilateral the product of the diagonals 
is equal to the sum of the products of the opposite sides. If a, 
b, c, and d are the sides, in order around the quadrilateral, and 
e and / the diagonals, ef=ac + bd; in the Greek trigonometry 
employing chords this theorem plays the same role that the 
formulas for sin (a + /?) and cos (a -\- /?) play in the trigonometry 
employing sines and cosines. A great French mathematician, 
Viete (1540-1603), the first to use generalized coefficients in 
algebraic equations, was the first to give these formulas, as 
sin (2 a + p) and cos (2 a + p) in terms of a + p and a ; the 



ADDITION FORMULAS 243 

modern form appeared hi 1748 in the work of the Swiss mathe- 
matician Euler. 

PROBLEMS 

1. Given « = 30°, £ = 45°, find sin(« + /3) and cos (« + /?). 
Check by tables. 

2. Given a = 60°, p = 45°, find sin 105°, and cos 105°. Check 
by the preceding problem, and explain the check. 

3. Given sin <* = -§, and sinp = T 5 -^, find sin (a + /J), when a 
and f$ are both acnte ; find sin (a + p) when a and ft are both 
obtuse ; when a is obtuse, /? acute. 

4. Given a and p acute angles, sin a = .351, cos p = .652, 
find sin (a + p) by the formula and check with the tables. 

5. Given sin 18° = .3090, cos 18° = .9511, find sin 36°. 

6. Given sin 18° = .3090, find sin 78°. 

7. Using the results of problem 1 for sin 75° and cos 75° 
with the data of problem 5, find sin 93° and cos 93° ; thus find 
sin 3° and cos 3°. 

8. What are sin (45° + a) and cos (45° + a) in terms of a ? 

9. Find sin (60° + a) in terms of sin a and cos a. 

6. The formulas for sin (a — 0) and cos (a — P). — If (3 is a 

negative angle, a — (3 comes directly under the a + (3 formula 
as «H-(-j8); if {3 is any positive angle greater than 360°, (3 
can be reduced to less than 360° by subtracting 360° (or some 
multiple of 360°) without affecting the functions of a — (3 or of 
/? : if (3 is positive and less than 360°, the functions of a — (3 
will be the same as the functions of a-|-(360 o — (3), since this 
simply adds one complete revolution to a — (3. Hence 
sin (a — (3) = 
sin (« + 360° - p) = sin a cos (300° - p) + cos a sin (360° - p) 

= sin a cos (— /?)+ cos a sin (— /?). 
cos (a - 0)=cos a cos (360° - p) - sin a sin (360° - /?) 

= cos « cos (— /?)— sin a sin (— 0). 



244 UNIFIED MATHEMATICS 

Substituting in these formulas, cos ( — ft) = cos ft, and 
sin (— ft) = — sin ft we obtain the subtraction formulas : 

sin (a — ft) = sin a cos ft — cos a sin ft, 
cos (a — ff)= cos ct cos /3 -f- sin a sin ft. 

7. Tangent formulas. — 

, / ■ , \ tan a + tan B . , m tan a — tan B 

tan (a -f #) = — £- ; tan (a — ft) = ^- • 

v y 1 — tan a tan ft v ^ 1 + tan a tan £ 

Since sin (a + /?) = sin cc cos /? -f cos a si 11 A 

and cos (« + /3) = cos a cos /? — sin a sin /?, 

for all angles a, and /?, without restriction, it follows that 

f ( I o\ _ sin (ct + /?) _ sin a cos /? + cos a sin /? 
cos (a + ft) cos a cos ft — sin a sin ft ' 

for all angles a and /?. 

Dividing numerator and denominator of the right-hand ex- 
pression by cos a cos ft, we have 



, / . m tan a + tan # 

tan (a + B)= — ^ 

1 — tan a tan / 



p 

. •-, -, , / ox tan a — tan /? 

Similarly, tan (a — ft) = - £- • 

J ' v ^ 1 + tan a tan £ 

8. Functions of double an angle. — The formulas for sin (a+ft), 
cos (a + ft), and tan (a + ft) hold if /? = a, whence 

sin (2 a) = sin (a + a) = sin a cos a -f- cos a sin a = 2 sin a cos a, 

sin (2 a) = 2 sin a cos a. 
Similarly, cos (2 a) = cos 2 a — sin 2 a = 2 cos 2 a — 1 = 1 — 2 sin 2 a. 
By division and simplification, or directly from tan (a + ft), 

, /r > \ 2 tan « 

tan (2 a) = 

K J 1- tan 2 a 

Note that whether a be regarded as positive or negative, i.e. 
as obtained by positive or negative rotation, as + a, or 
-f a — 360°, 2 a has the same terminal line as 2 a — 720°. 



ADDITION FORMULAS 245 

PROBLEMS 

Note. — See the preceding list of problems, and use numerical values 
as There computed. 

1. Given a = 45°, ft = 30°, find sin (a - ft) and cos (« - ft), 
checking by the tables. 

2. Given a = 60°, ft = 45°, find sin (a — ft) and compare 
with problem 1. Find tan (a -f- ft), tan (a — ft), and tan 2 a. 

3. Given sin a = f and sin/3 = y 5 3, find sin (a — ft) when 
a and /? are acute ; find sin (a — ft) when a and ft are both 
obtuse. Explain the result ; find sin (a — ft) and cos (a — ft) 
when a is obtuse and (3 is acute. Interpret. Find 
tan (a — ft), tan (a -\- (3), and tan 2 a for a and j3 in I. 

4. Given a and /? acute, sin a = .351 and cos (3 = .652, find 
sin (a — (3) and check by the tables. Find tan (a — (3). 

5. Given sin 18° = .3090, cos 18° = .9511, and sin 15° from 
problem 1, find sin 3° and cos 3°. 

6. Find sin 42° as sin (60° - 18°). 

7. Express sin (60° — a) and cos (60° — a) in terms of func- 
tions of a. 

8. Find the value of sin (60° + a)— sin (60° - a). 

9. Find the value of cos (45° + a)-f cos (45° — a). 

10. Show that sin (a + (3) sin (a — f$)= sin 2 a — sin 2 ft. 

11. Find a value of cos (a + /8) cos (a — /?), similar to the 
pre ceding. 

12. Given tan a = 1.4, find tan 2 «. 

13. Given cos 2 a = .63, find sin a and cos a ; are there two 
solutions ? 

14. Given that one line cuts the sc-axis at an angle a such 
that tan a = 3, and another line cuts the a>axis at an angle f3 
such that tan ft = -J-, find the tangent of the angle between the 
two lines by assuming that they intersect on the coaxis. 
Check by using the tables to find the slope angles of these 
lines. 



246 



UNIFIED MATHEMATICS 



9. The tangent of the angle between two lines. — Given any 
two lines as y = 3x — .5, y — — x — 7, it is evident by plane 
geometry that the angle between them is the same as the angle 

between y = 3 x, y = — x, 
lines parallel to these 
given lines through the 
origin. The word "be- 
tween" implies no dis- 
tinction as to priority of 
either line ; thus the 
angle may be taken as 
either a positive or nega- 
tive acute angle, or the 
corresponding supple- 
mentary angle. Thus if 
the lines were inclined 
to each other at 30°, 
the angle might be con- 
sidered as + 30°, - 30°, 
+ 150°, -or -150°; the 
tangent of the angle 
would then have the 




Angle between two lines 



the same angle. 
value given by the expression ± 



V3 



To distinguish between the two lines we may say that we 
wish the angle from the line of slope -f 3 to the line of slope 
— 1, or in the general case, from 
the line of slope m 2 to the line of 
slope m x ; by analogy with our 
use in defining the angle which 
a line makes with the as-axis, when 
we say the angle which the line 
y — — x makes with y = 3x we 

mean the angle obtained by re-. Angle between two lines 

volving the line whose slope is <p = e x - e 2 , &i > 6 2 . 




ADDITION FORMULAS 



247 



3 so as to make it coincide with the line whose slope is — 1. 
Calling the angle whose tangent is 3 (written tan" 1 3 or arc 
tan 3, meaning the angle whose tangent is 3), 2 , and the angle 
whose tangent is — 1, e 1} we find that the angle <£ from the $ 2 
line to the 6 X line is <z> = l — 6 2 . 

tan <f> = tan (0, - fl 2 )= tan^-taii^ = m l - m 2 < 
1 -f- tan 1 tan 2 1 + rn^nx^ 

If the two lines are parallel the angle is 0, hence tan <£ = 0, 
and m± — m 2 = 0, or m 1 = m 2 , as anticipated ; if the lines are 
perpendicular tan</> becomes infinitely large, and for finite 
values of m x and m 2 (excluding lines parallel to the axes), the 



denominator 1 + ??i 1 m 2 = 0, or m 2 = — 



i.e. 



nu 



the slope of a 



perpendicular is the negative reciprocal of the slope of the 
given line. When one line is parallel to the y-axis, its slope 
m 2 (or mi) is infinite, but the angle between the two lines can 
be obtained by dividing numerator and denominator of tan <£ 



by m 2 (or m{), giving tan <f> 



nh 1 



1 J ' 

hwzi 

m 2 



or 



— when m 9 ap- 



proaches infinity, for the tangent 
of the angle made by a given line 
with the i/-axis. 

tan <f> = —^ L gi ves the an _ 

1 + 777x7722 

gle from the m 2 line to the m x 
-line. 

If 2 > h <f> is negative, but 
the formula <f> = $ 1 — 6 2 still holds. 




10. Functions of half an angle, cos (2 a) = cos 2 a 
all values of a. 

Substitute x for 2 a, and - for a. 



sin 2 a for 



248 



UNIFIED MATHEMATICS 



wmmmm 


Mliii 


Ijj- L44- 


_ ^. J_ 

-- 2- 



cosa = cos 2 - — sin 2 -• 
2 2 

l = cos 2 - + sin 2 -. 
2 2' 



1 -f cos x = 2 cos 2 
whence 



2' 



Half-angle relations 

Similarly, 



cos - = ± V|(l + cos x). 



sin? 
2 



± V|(l — COS X). 



tan ^ = ± Jl- CQS g = ± J O- ~ cos ^X 1 - CQS x ) , 
2 *l + cosx * 1 — cos 2 # 

+ if ic is in I or II, and — if x is in III or IV ; the formula 

x 1 — cos x 



tan - = 
9 



is one in which 



sinx 



the sin x takes care of the algebraic sign ; and so also 

x sin x 



tan 



by rationalization. 



2 1 + cos x' 



both 



Note that if x is regarded as a positive angle, less than 360°, 

x 
sin - is + ; but the same position of the terminal line 
it 

is obtained by x ± 360° ; | and ^+180° have sine and 

Z Z 



cosine opposite in sign, but tan ( - ± 180 c 



tan 



Since 



cos (— a;)= cos (x) it must be stated whether x is in I or IV, or 
in II or III ; i.e. cos x alone does not locate the angle x. 

If # in I is regarded as a positive angle, — is -f- acute, and 

Z 

sin - and cos - are positive ; if # in I is regarded as a negative 

Z Z 

reflex angle, - is negative obtuse, and sin - and cos - are both 

Z Z Z 



ADDITION FORMULAS 249 



negative ; in either ease tan - is positive. Similarly if x is 

taken in II, III, or IV, the formula takes care of all positions, 
proper account being taken of the algebraic sign of the radicals. 



PROBLEMS 

Find the angle between the following lines : 

1. y = 3 x — 5, and y = — x — 7. 

2. y = 3 x— 5, and y = x — 7. 

3. 2 y - 3 x - 7 = 0, and 3 y + 4 x -5 = 0. 

4. 3 ?/ = 5 x — 5, and ?/ = 8 x — 10. 

5. 3y = ox — 5, x = 5. 

6. 3# = 5a— 7, 2/ = 5. 

7. In the preceding 6 problems, find the tangent of the angle 
made by the first line with the second line, i.e. the tangent of 
the angle obtained by rotating the second line until it coin- 
cides with the first. Why is it that the sense of this rotation 
is immaterial ? 

8. In the above problems check by finding from the tables 
the trigonometric angles involved. 

9. Find the pencil of parallel lines making an angle of 30° 
with each of the lines in problem 1 ; find the one of the family 
through (-3,5). 

10. Find the pencil of lines making an angle of 45° with 
each of the lines in problem 3 ; find the particular one 
through the origin. 

11. Find the pencil of lines making an angle of 90° with 
each of the lines in problem 4. 

12. Given sin 30° = .5000, cos 30°= .8660, and tan 30° =.5774, 
find sin 15°, cos" 15°, and tan 15°. 

13. Find sin 7-J- , cos 7-1°, tan 7-±°, using half-angle formulas. 

14. Given sin 45° = cos 45° = .7071, find sin 221°, cos 221°, 
and tan 221°. 



250 UNIFIED MATHEMATICS 

15. Use sin (a — /?) formula to obtain sin 7-J- and cos 7^°, 
from the functions of 30° and 22|°. Compare with problem 13. 

16. Given sin 18° = .3090 and cos 18° = .9511, find sin 12° 
and cos 12°. 

17. From the functions of 12°, compute the functions of 6°, 
and then the functions of 3° and of l-J-°, using half-angle 
formulas. 

18. Compute sin 1^° and cos 11° by the difference formulas- 
taking li° as 7i°- 6°. 

19. Compute the functions of -§° from the functions of 1°. 

20. Find by interpolation sin 1° and cos 1° from the com- 
puted values of the functions of |° and 1± . Compare with 
the tabular values. 

21. Make a table of values of the sine, from to 45° in- 
creasing by 1|° intervals. 



CHAPTER XVI 



TRIGONOMETRIC FORMULAS FOR OBLIQUE 
TRIANGLES 

1. General statement. — Employing elementary theorems of 
plane geometry it is possible to construct any triangle when 
given the three sides, or two sides and an angle, or one of the 
three sides together with two of the angles ; in trigonometry 
the corresponding problem is the numerical solution, not sim- 
ply the graphical, of the types of triangles mentioned. The 
trigonometric solution which has been given of the different 
types of right triangles, with unknown parts, can be applied 
to effect the trigonometric solution of any oblique triangle ; 
but in general, these methods do not give convenient formulas 
for computation. As the general triangle is fundamental in 
surveying (note the term " triangulation "), in astronomical 
work, and in many problems in physics, more convenient 
formulas than those given by right triangles are a practical 
necessity. 

In general the laws and formulas of plane trigonometry 
connect directly with proposi- 
tions of plane geometry ; the 
effort is to express the inter- 
dependence of the angles and 
sides in the form of equations 
involving the trigonometric func- 
tions of the angles. 

The vertices of any triangle 
being lettered A, B, C, it is con- 
venient to designate the corresponding angles at these ver- 
tices by a, (3, and y, respectively, or by A, B, and C, if no 

251 











( 




i i 


- T i :: 






J_L 


i i 




























iiVv- 


TtnJ ! i 


_ — | __ 




Bff 

A 


LsH±; 

a 

h — —0-- 


3|§|::: 



A, B, C; a, p, y; a, b, c 



252 



UNIFIED MATHEMATICS 



confusion of meaning is possible ; the sides opposite A, B, and 
C are designated by a, b, and c, respectively. 

2. Cosine law. — If the two sides of a triangle are given, the 
third or variable side, opposite the angle a, between the two 




a 2 = b 2 + c 2 - 2 be cos a 

given sides, evidently changes as a changes. Let b and c 
remain fixed. Let M be the foot of the perpendicular from 
C upon AB ; then AM = b cos a, for any angle a when the 
direction AB is taken as positive. Further in every position 

MB = AB - AM= c - b cos a, 
for in every position AM -\- MB == AB. 
The altitude ikfO = h = b sin cc. 
Hence, .BO 2 = MB 2 + J/C 2 

= (c — b cos a) 2 -f- (b sin a) 2 
= c 2 — 2 &c cos a + 5 2 (cos 2 a + sin 2 a). 
a 2 = 6 2 -|- c 2 - 2 6c cos a. 

All limitations upon a are removed by 
the different types of figures. Hence 
for any angle a, 

^^S|===|ffi « 2 = & 2 + c 2 - 2 be cos a 

gives the length of the side a, opposite 
a, in terms of the other two sides and a. 
Since a and a may represent any side 
and the opposite angle of any given tri- 
angle, b and c being the other two sides, our formula may be 
stated as follows : 




M falling outside B 
Formula unchanged 



FORMULAS FOR OBLIQUE TRIANGLES 253 



The square of any side of a triangle is equal to the sum of the 
squares of the other tivo sides less twice their product into the 
cosine of the including angle. 

Or, TJie cosine of any angle equals the difference between the 
sum of the squares of the tico including sides and the square of 
the side opposite, divided by twice the product of the including 
sides. 

If a, b, and c are the sides of any triangle, with a, /?, and y 
the corresponding opposite angles, we have the ' following 
relationships : ft2 = &2 + & _ 2 u cos ^ 

b 2 = c 2 + a 2 — 2 ac cos 0, 
c 2 = a 2 -f b 2 — 2 ab cos y ; 

62 _|_ C 2 _ a 2 



or 



cos a = 



cos /? = 



cos y 



2&c 
c 2 + a 2 -b 2 

2ac 
a 2 + y - c 2 

2 aft 



3. Cyclic interchange. — Any formula which has been de- 
rived, without imposing any limitations upon a, b, c, a, /3, or y, 
connecting a, b, c, and trigono- 
metric functions of the angles a, 
ft, and y, will continue to hold if 
a and b and, at the same time a 



and /?, are interchanged ; or if 
b 



b 



are changed to 

such changes effect 



a to r . 



I a 
[a 

and 



to 





^--1 ■-- 


4k£ SS\ 


ii/y N\K a 


jZ/» £a 




I i 1 




A\V Ut 


-CS ^4^ 




\N<^ ~r— ' "\, xj/ 


^-c-^ 


^ 



Cyclic interchange 



iy 1" 

simply a re-lettering of the figure. 

The change of a into b, b into c, 

and c into a is called a cyclic interchange of the letters a, b, 

and c. Note that cyclic interchange gives the second formula 

from the firsts and the third from the second. 



254 UNIFIED MATHEMATICS 

In the figures, a in the first is chosen as an acute angle, but 
this limitation is removed by deriving the same formulas for 
a a right angle and for a an obtuse angle ; c is taken as 
longer than b, but interchanging b and c in our derived for- 
mula leaves the formula unchanged ; assuming b and c equal 
would involve no change whatever in our proof ; and if b 
is assumed greater than c, a fourth figure can be drawn in 
which M falls beyond B on AB produced ; but the formula 
a 2 = b 2 + c 2 — 2 be cos a remains the same, as the student may 
easily verify. 

PROBLEMS 

In the following problems use .866, .707, and .500 for the cosines of 30°, 
45°, and 60°, respectively. 

1. Given b = 140, c = 230, a = 60°, compute a. Refer back 
to the section on extraction of square root, page 23. 

2. Compute a when a = 30°, and 45°, 90°, 120°, 135°, 180°, 
when b = 140, c = 230. 

3. Given a = 155, c = 234, = 35°, compute b. What 
changes in b are effected by changes of ± 10' in (3 ? 

4. Given a = 155, c = 234, compute (3 when b = 172. What 
is the maximum change in ft which an error of ± -J- unit in a, 
b, and c could introduce, (3 being computed to minutes? Take 
1551 2341 w ith 1711 ; take 1541 and 2341, w ith 1721 Note 
that (1551) 2 and (1541) 2 differ from (155) 2 by about 155 ; 
similarly with the other values ; if the squares are found by 
logarithms it is well to look up log 155.5 and log 154.5 at the 
same time as log 155, etc. 

5. In problem 1, find cos /?, and then /?, taking for a the 
value obtained there. 

6. In problem 1, find cos y and y, using the computed value 
of a. Check by summing ft and y with the given angle. 

7. Given a =200, 6 = 150, c = 300, find a. What change 
in a would a change of ± 1 in a effect ? Suppose that a, b, 



FORMULAS FOR OBLIQUE TRIANGLES 255 

and c are given only to two significant figures, i.e. a is between 
195 and 205, b is between 145 and 155, c is between 295 and 
305, compute a and discuss limiting values. 

8. Compute the third side in the following 5 problems, using 
logarithms for squaring ; time yourself in the exercise. Fifty 
minutes should be ample time for the 5 problems ; devise a 
convenient form and use it in each example. 

a. Given a = 366, b = 677, y = 15° 10'. 

b. Given a = 423, c = 288, £ = 35° 15'. 
a Given b = 627, c = 816, a = 100° 41'. 

d. Given a = 635, c = 341, p = 67° 38'. 

e. Given c = 184, b = 295, a = 130° 54'. 

4. Sine law. — A circle may be circumscribed about any 
triangle ; let the radius be designated by R. The figure shows 




Sine law : 



sin a sin p sin y 



that if A is an acute angle, sin a = ^ = — ; if a is 90°, this 

R 2 R 

formula is still true, as a equals 2 R, and the formula gives 

sin 90° = 1 ; if a is obtuse, the figure gives sin (180° - a) — — , 

a ^ ^ 

whence sin a = . 

2R 



256 



UNIFIED MATHEMATICS 



Therefore without any limitation whatever upon a, 

















_± s _ % _ _ _ 




-j_ * », 




T/ s 




4v \ 




J- 1 




- jE - - s 




Xtt- 3 




^'dr 








rP^P""* t 








~ 5u "^ :""--z;: ~ ~i ~ 




|4/j s * *■"— i 












\\ * I *•-. <-*i'i\ 




\\\ N 1 >VT // 




V \ 6* ^7 




M f J h-a'p 












s v 5* 
















1 1 



sm a = 



2iT 

interchange of letters gives 
b 



sm f$ = 



and 

Whence 
a 



sm y 



2i?' 

c 
27?' 



2R. 



Sine law : a obtuse 



sin a sin /? sin y 
This formula states that in any 



triangle the ratio of the side op- 
posite any angle to the sine of that angle is constant, and this 
ratio is numerically equal to the diameter of the circumscribed 
circle. 

Further, = " = 2 or the ratio of the sine of any 

a b c J 

angle to the side opposite is constant. 

Note that if 2 R is regarded as the chord "of 180° of the 
circle in which the triangle ABC is inscribed, the proposition 
states, in effect, that in any circle any chord is proportional 
to the sine of the inscribed angle which intercepts the arc of 
the chord. 

The formula may be stated : 



sm a 
a 



sin/? 
b 



sin y sin 90° sin 30 c 



2- 



sin k c 



2R 



R 



chord (2k°y 

all of the chords being chords of the circle circumscribed about 
the triangle. The ratio of the sine of any central angle in a 
circle to the chord of double the angle can readily be shown to 
1 



be constant, 



2R 



5. The sine law historically. — The sine law was discovered 
by an Arabic (Persian, by birth) mathematician, Nasir al-Din, 



FORMULAS FOR OBLIQUE TRIANGLES 257 

at-Tusi, who lived 1201-1274 a.d. To him we owe the first 
systematic treatise on plane trigonometry, an achievement made 
possible by the combination of the Greek trigonometry using 
chords with the Hindu trigonometry employing sines. To 
Europeans the sine law was communicated by the great Ger- 
man mathematician and astronomer, Regiomontanus, in his 
work on trigonometry, De Triangulis, the first published sys- 
tematic treatise ; it was published at Nuremberg in 1539, many 
years after the death of Regiomontanus, who lived 1436-1476. 

PROBLEMS 

1. Given a = 150, b = 200, a = 30°, find sin p using natural 
functions, 

2. Given a = 150, a = 30°, p = 45°, find b, using natural 
functions. 

3. Given a = 150.4, b = 214.3, a = 31° 10', find sin p employ- 
ing logarithms. 

4. Given a = 150.4, a = 31° 10', p = 44° 16', find b by loga- 
rithmic computation. 

5. In the formula, a 2 = b 2 -f- c 2 — 2 be cos a, substitute the 
values as given in problem 1 and solve for c. Note that there 
are two solutions. What is the explanation ? 

6. Time yourself in solving the following set of 6 problems, 
applying the sine law ; make a type form of solution and use 
it in each problem. Thirty minutes should be sufficient for 
the 6 problems. 

a. Given a = 366, b = 677, a = 15° 10'. Find sin /3 and /?. 

b. Given a = 423, c = 288, y = 35° 15'. Find sin a and a. 

c. Given a = 627, a = 100° 11', /? = 43° 15'. Find b. 

d. Given b = 816, (3 = 67° 18', y = 34° 9'. Find c. 

e. Given c = 635, p = 130° 14', a = 20° 12'. Find b. 
/. Given b = 284, « = 40° 10', p = 35° 15'. Find c. 



258 



UNIFIED MATHEMATICS 



6. Half-angle formulas. — As the circumscribed circle has 
yielded a formula of great value trigonoinetrically the in- 
scribed circle may be examined trigonometrically with the hope 

of a similar result. 

The bisectors of 
the three angles of 
the triangle meet in 
a point which is the 
center of the in- 
scribed circle ; let 
this circle be drawn 
and let L, M, N, be 
the points of tan- 
gency, then 

AM=AN, 
BL = BN, 
OL = OM = ON = r, radius inscribed circle CL = CM, 

being tangents from an exterior point. Evidently the six 
segments mentioned make the perimeter, 2 s, of our triangle ; 
2s = a-\-b-\-c\ adding above we have that 

AM + BL + CL = AN+ BN+ CM= s, 

but BL-\-CL = a, and BN+ AN= c ; 

whence AM = s — a, CM= s — c, and similarly 

BN=BL = s-b 

ON=OL=OM=r. 




tan a = ~ 

2 s 



-; tan-0 

a 2 



-b 



tan- r = 



s — c 



7. Area. — In the preceding section the area of the given 
triangle is easily determined in terms of r and s, for the area 
equals the sum of the three triangles on a, b, and c as bases, 
each having the altitude r. .*. A = ^ r(a + b + c) = rs. 



FORMULAS FOR OBLIQUE TRIANGLES 259 

However, if the three sides are given, this formula does not 
enable us to determine r without using a further formula to 
determine A. 

A = J be sin a = ±ac sin ft = J a6 sin y. 

This type of formula for the area is applicable when two sides 
and the included angle of a triangle are given or found. 

yi _i_ c 2 a 2 

A—\ be sin a can be combined with cos a= — — in such 

2 be 

a way as to eliminate a, giving A in terms of the three sides. 
A 2 = ± b 2 c 2 sin 2 a = ± 6 2 c 2 (1 - cos 2 a) 

— 1 &2 C 2 (1 _ cos a )(l + CO g a ) 



bW ± 


&2 +c 2_ a 2\/ ^2 + c 2_ a 2> 

2 6c- A 2 6c , 


6 2 c 2 /Yt 2 - 
" 4\ 


- 6 2 + 2 6c - c 2 V6 2 + 2 6c + c 2 - 
2 6c J\ 2 6c 


_ 6 2 c 2 [a 2 - 


-(6-c) 2 ][(6 + c) 2 -a 2 ] 




16 6 2 c 2 



_ (a— 6 + c)(a + 6 — c)(6 + c— a)(b + e + a) 
2.2-2.2* 

But a + b + c = s,zna a - b+c = a + b + c -b=:s-b,etc. 

2 2 2 

The above formula for A 2 may be written, 
A 2 = s(s-a)(s-b)(s-c) 
A = Vs(s — a)(s — 6)(s — c). 



Further J. = rs, whence r = — = aM 



)(s-6)Q-c) 



This value of r is employed with the half-angle tangent 
formulas of the preceding article to determine the angles of a 
triangle when the three sides are given. 



260 



UNIFIED MATHEMATICS 



8. Newton's check formula. — A formula which involves all 
of the sides and all of the angles of an oblique triangle is 
particularly desirable as a check formula to be used upon the 
results obtained by direct application of the sine law or in a solu- 
tion obtained by right triangles. Such a formula was devised 
by Sir Isaac Newton and appeared in his ArithmeMca universalis 
of 1707 ; our proof follows the lines of that by Newton. 

































































































































































































F 
















































A 
















































£\ 
















































J? ^ 
















































X i 














































£_ 


-t^J- 














































■' 


-sb-t 












































^ 




l_ 












































s 














































S 
















































rC* 






l_ 










































k,' 






















































l_ 








































/' 


4 V 






















































f - 








































£ 








"1 








































Kt> 








/ 








































<«.> 


tyrt-\ 






T 








































1 




\ 




- £ J 








































1 




\ 




. (L 








































t 




N . 


ft"/ 


%)i- 








































1 




\%T" 




£ 








































1 


















































v-^/^S 


_/. 








































4 


<; 


b i / 




./ 
















\ 


P 






















_/_ 




U4° 


1 \ 










































_i 




11 




v 


























































jL> 






























Mi 








A 



















































































































































Let ABC be any triangle ; 

from C draw the bisector CE of the angle ACB or y ; 

extend BC to .F, making CF=CA = b\ 

AF is parallel to C.2J, by plane geometry ; 

angle CFA = angle BCE = | y. 

Now angle £AF = a + -J- y = 90° - i(a - /?),' 

since i« + |i8 + h = 90 °- 

Applying the sine law to the triangle BAF, we have the 
desired formula : 

a + & __ cos |(q — p) 
c sin -J- 7 



FORMULAS FOR OBLIQUE TRIANGLES 261 



By drawing the bisector of the exterior angle, and drawing 
a parallel from A, a second useful check formula is obtained : 
a — b _ sin ^(q — p) 
c cos ^ -y 

Noting that y y = 90° — ^(a -f /3), division of the second 
equation by the Xewtoniau, member for member, gives 
a- & ^ tani(q-[3) . 
a + b tan ± (a + P) ' 
this symmetrical formula is known as the tangent law. 

Cyclical interchange gives in each one of the above two 
corresponding formulas. 

9. Historical note. — The formula A = Vs(s — a)(s — b)(s — c) 
was first given by Hero of Alexandria, first century a.d., 
a teacher of mathematics and mechanics in what was probably 
a kind of technical school at Alexandria in Egypt : it is called 
Hero's formula. 

An extension of this formula is given by Bhaskara, a Hindu 
mathematician of about 1000 a.d. Bhaskara's formula gives 
the area of any quadrilateral which is inscribable in a circle, 
i.e. with the opposite angles supplementary, as 



A = ^/(s — a)(s — b)(s — c)(s — d). 
The triangle may be regarded as a special case with d = 0. 

10. Reflection and refraction of light. — Rays of light, like rays 
of heat and sound and elec- 
tric rays of various types, 
travel in straight lines from 
the source. Rays of light 
emanating from the sun 
travel in nearly parallel 
rays, since the point of con- 
vergence, the source at the 
sun, is at so great a dis- Reflected ray travels shortest path 
tance from the earth. A ray of light which meets a polished 
plane surface, a mirror, is reflected at an angle which is such 



\T 


X M- - X 


X x _ _ _± 




/ 


! ^" 


1 ' ' ' -,"* 


!!; i ; ; : ! ^' 






"^-^ x x ^^ ' 










^-S ^- l w" 








: Ui >^~-~" . J- - 4- 


IT — — T 







262 



UNIFIED MATHEMATICS 



as to make the total path from the source (L) to the reflect- 
ing surface and then to a second position (R) the shortest 
possible. LSR is the shortest distance from L to 8 to R if 
the angle of incidence i, made by the original ray with the 
normal, to the surface at S where the ray strikes, is equal to 
the angle of reflection r. Evidently LSR = LSR' ; the straight 
line joining L to R', a point symmetrically situated to R with 
respect to the polished surface, is shorter than any other line, 
for any other broken line LS'R = LS'R' is greater than the 
straight line LSR' and hence greater than LSR. 

If the ray of light meets, not a polished surface but some 
transparent medium, other than that in which the ray is 
traveling, the ray of light is not continued in the same 
straight line in which it starts but it is broken, or refracted, 
continuing on its path in a straight line which makes a differ- 
ent angle with the normal than does the original, incident ray. 

It is found by physical ex- 
periments that the angle of 
refraction, the angle of the 
refracted ray with the nor- 
mal, bears a simple relation 
to the angle of incidence, 



"' ' la 1 


Jyr + 












_ _ -_ it- m *Z 


\- -%* 


\** 


** 


±1_ _::_:::::::::::: 




« ? 2: £l 


^* ? i - 


2 2 - 


*£,*£ -2Z . _ _ 


T2? iZ^Z - 












ft- J L --4 v h 



sin i 
sinr 



= 7c, wherein k depends 



Refracted ray of light 



upon the nature of the two 
media through which the 
light is passing. Thus for a ray of light passing from air, a 
rarer light medium, to the denser water the value of k is J, 
sin i _ 4 
sinr 3 

A student who thoughtfully examines this formula will be 
reminded of the sine law, which does indeed give a very simple 
construction for the refracted ray when the constant k is known. 

Let two concentric circles be drawn whose radii bear to 
each other the ratio, J ? f the index of refraction. In the 



FORMULAS FOR OBLIQUE TRIANGLES 263 

figure the ratio is taken j, the index of refraction for light 
from air to water. Extend LO, the incident ray, to L', cutting 
the circle of smaller radius. From V drop a line parallel to 
the normal N'O to cut the larger circle in R. Connecting R 




LO is the incident ray ; OR is the refracted ray 

with gives the refracted ray. In the triangle OL'R, the 
Z OL'R = 180° - i, and the Z ORV = Z r, of refraction ; by 
the sine law 

sm(180°-i) ^sin^ = 4 
sin r sin r 3 



From water to air the index is f, it being found that if the 
refracted ray is replaced by an original ray, this new ray 
in the second medium will be refracted along OL, the path of 
the incident ray with which we started. 

The construction for the refracted ray in air for a ray of 
light emanating from the water, RO, is entirely similar to the 
preceding. RO is extended to R' on the larger circle. From 



264 



UNIFIED MATHEMATICS 



R' a parallel B'L is drawn to the normal to cut the smaller 

circle. OL is the refracted ray. 

Evidently if sin r = f sin i, sin r is 
always less than f . If a ray of light 
starts from any point within the arc 
A'T wherein T is the intersection of 
the vertical tangent to the smaller 
circle with the larger circle it cannot 
be refracted into the air at 0, and the 
whole light is reflected at 0. This 
property of the light rays is utilized 
in certain spectroscopic work. Thus in 
the case of a glass prism, index of re- 
fraction f, if light strikes the plane sur- 
face at an angle of incidence greater than 
41° 48', since sin 41° 48'= .6666, or f, 
all the light will be reflected ; this type 




Glass reversing prism 
Index of refraction f 

Angle of incidence, 45°. 
For any angle i greater 
than ct, sin a = f , the 
beam is reflected. 



of prism is used in projecting lanterns. 



PROBLEMS 

1. Given a = 9, b = 14, c = 19, find the area of the triangle, 
using Hero's formula. 

2. Given a = 9, b = 14, c = 19, find a, using the cosine law. 

3. Given a = 9, b = 14, c = 19, find the area by the formula 
A — \ be sin a. 

4. Given a = 9.34, b = 14.31, c = 19.27, find the area by 
Hero's formula, using logarithms. 

5. Given a = 9.34, b = 14.31, c = 19.27, find a by the cosine 
law, and then find the area using the formula involving sin a. 

6. In the two triangles above find r, the radius of the in- 
scribed circle, using r • s = A. 

7. In the two angles above find tan i a, tan \ /?, and tan \ y, 
using the half-angle formulas. Eind the angle sum in each 
case. 



FORMULAS FOR OBLIQUE TRIANGLES 265 

8. Draw circles with radii two inches and three inches and 
show how to construct the refracted rays of light passing from 
air into glass at angles of incidence of 30°, 45°, and 60°. 

9. For what angle will a ray of light passing from glass 
into water be reflected, and not refracted ? The index of re- 
fraction of light passing from glass into water is J-. Draw the 
figure. 

10. Find the angle of refraction of rays of light passing from 
air into water, fc = 1 .33, when the angles of incidence are 
31° 15', 37° 18', 44° 25', 67° 10', 83° 15'. For which of these 
angles is the course of the ray changed by the greatest amount ? 

11. Suppose the rays in problem 10 to pass from air into 
glass, solve for the angles of refraction. 

12. Construct two of the figures in both problems 10 and 
11, and check graphically the results obtained above. 



CHAPTER XVII 
SOLUTION OF TRIANGLES 

1. Solution of triangles given two angles and one side : cap 
type. — With surveying instruments the simplest method of 
locating the distances from two fixed points to a third inacces- 
sible point is to determine the length AB and the angles a and 
(3, at A and B respectively, wherein A and B are two points 
from which C is visible. Using the sine law, 

a __ b c 

sin a sin /3 siny' 

a c 

we select the equation 



sm a 



sin y 



b c • • 

or =— — , since in each of these only one unknown 

sin (3 sm y 

quantity appears. The third equation a = , not in- 

sin a sin ft 

dependent of the other two, is used as a partial check upon 

the computed values. As a more complete check use Newton's 

formula ' a + & _ cos -!(<*-/?) 

c sin i y 

This form of triangle appears in the classical problem, whose 
solution by plane geometry is ascribed to one of the seven wise 
men of Greece, Thales of Miletus, sixth century b.c. The prob- 
lem is familiar to the surveyors, being used in determining 
distances across a stream, or to an inaccessible point. The 
astronomer has the same problem in locating the distance of 
fixed stars using two observations, at different points in the 
earth's orbit, of the angle made by lines from the earth to the 

266 



SOLUTION OF TRIANGLES 



267 



star and to the sun ; for simplicity, the two points of observa- 
tion may be considered as taken at the extremities of the 
diameter of the earth's path. 

In locating batteries by the sound waves this type of 
triangle is employed ; two or three observers at different 
points can locate an enemy battery .by this method within a 
radius of fifty feet or thereabouts. 

2. Type form of solution : caf type. — The form of the solu- 
tion is important ; follow the given form closely. 

Given a = 65° 11', £ = 38° 24', c = 175 feet. Find a and b. 



c sm a 



(written from the formula 



which 



sm y 
should not be set down). 

6= csin_g, cheek6= asm/3 
sm y sin a 

A 1, • 1 c 2 since sin /3 
A = -bc sma= = K. 

2 2 sm y 

a = 65° 11' 
P = 38° 24' 
y = 76° 25' 

logc = 2.2430 

+ log sin a = 9.9580 - 10 



sm a 



sin 



12.2010 - 10 
-log sin y= 9.9876 - 10 
log a = 2.2134 

a = 163.4 

log a = 2.2134 

+ log sin = 9.7932 - 10 

12.0066 - 10 

-log sin «= 9.9580 - 10 

log b = 2.0486 - 10 

But log b = 2.0486 by above 

computation, which checks. 




Two angles and a side given 

logc = 2.2430 

+ log sin = 9.7932 - 10 
12.0362 - 10 

- log sin y = 9.9876 - 10 
log b = 2.0486 

.6 = 111.8 

logc 2 = 4.4860 

+ log sin a = 9.9580 - 10 

+ log sin p = 9.7932 - 10 
14.2372 - 10 

- log sin y = 9.9876 - 10 
log 2 A = 4.2496 

2A= 17,760 
A= 8880 



268 UNIFIED MATHEMATICS 

The check which we have used is only partial as an error in 
y or sin y would be carried through the work without showing 
up in the check. The Newtonian formula gives a real check 
upon the computation. 

Check. g + 6_Q08K«-fl 





c sin i y 


- h = 275.2 


log (a + 6) = 2.4396 




log c = 2.2430 




.1966 


£ = 26°47 r 


log cos i («-£) = 9.9880 


y = 76° 25' 


log sin |y = 9.7914 



.1966 which checks. 

Notes. — The whole form of solution is placed on paper before the 
logarithms are inserted. Place the given angles in vertical column and 
obtain the third angle by noting the angle which added to the given 
angles makes 180° ; thus, here note first that to complete 11' and 24' to 
1° takes 25'. Add this 1° to the 8° and 5°, the units of our given angles, 
making 14° ; complete by 6°, which is written in its proper place, to 20°. 
Carry the 2 tens, to the tens, making 11 tens, or 110°, .requiring 7 tens 
(written in the proper place) to complete to 180°. 

Look up logc, i.e., log 175, writing this immediately in all places 
where it occurs ; for the area, it is simpler to calculate 2 A and divide by 
two than to divide by subtracting log 2 in the work. log c 2 = 2 log c, 
which is set down in its place. Finish, as far as possible, with the logs 
of numbers before taking up the logs of trigonometric functions, 
log sin 65° 11', log sin 38° 24', and log sin 76° 25' should be found in the 
order in which they occur in the tables, to avoid useless thumbing back 
and forth ; any value found should be immediately inserted wherever 
it occurs in the form. 

PROBLEMS 

1. Prove the sine law by using perpendiculars dropped from 
a vertex to the opposite side. 

2. Given c = 350.4, a = 36° 14', /3 = 100° 24', find b and a, 
by the sine law. 

3. Given a = .03504, a = 36° 14', /3 = 100° 24', find b and c, 
by the sine law. 



SOLUTION OF TRIANGLES 



269 



4. Solve completely the following 5 triangles ; take the 
rime of your solutions ; write the complete form of solution 
for each problem, in turn, before inserting any logarithms. 
The five problems should be completed within one hour and 
20 minutes using the rough check by the sine law. As a 
separate exercise check all by Newton's formula, timing 
yourself. 

a. a = 627 a = 100° 11' 

6. '6 = 816 /3=67°18 r 

c. c = 635 (3= 130° 14' 

d. 6 = 284 a = 40° 10' 

e. a = 366 ■« = 15° 10' 



= 43° 15' 
y = 34° 09' 
a = 20° 12' 
P = 35° 15' 
= 95° 14' 



3. Given two sides and the angle opposite one : aba type. — 

Given b, a, and a to construct the triangle geometrically. AC 
is laid off of length b and the line AX is drawn so as to make 
Z CAX = a. Since a must lie opposite to a, a is taken as 



■ 

— 


1 1 ' ' 


§ 


! i 


1 1 1 lr 

SHii 

zttgtJH::: :::: 


g 

= = ==== j 


3k === == = === 


m 


1— 


s= 


441 

i i 


tt iSf 

-h4h — i — +4 


u======% 


IBM 



Given two sides and the angle opposite one 

The side opposite the given angle must always be greater than, or equal 
to, the corresponding altitude. 

radius and with C as center an arc is swung to cut the side 
AX. Since the shortest distance from C to AX is the length 
of the perpendicular CM, if a is given less than this perpen- 
dicular there is no solution. If a is given equal to the per- 
pendicular there is one solution ; if a is greater than the 
perpendicular the arc cuts AX in two points, but unless a < b 
the one point of intersection to the left of A will not repre- 
sent a solution. The perpendicular is of length b sin a ; if a 



270 UNIFIED MATHEMATICS 

is equal to or greater than 90°, there will be one solution if 
a > b, and none if a <^b, for the greater angle lies opposite 
the greater side. By plane geometry then, we have the fol- 
lowing scheme, indicating whether one solution, two solutions, 
or no solutions are possible. 

a > 90°, a j< b, no solution. 

a > 90°, a > b, one solution. 

a < 90°, a < b sin a, no solution. 

a < 90°, a = b sin a, one solution. 

a < 90°, b sin a < a < b, two solutions. 

a < 90°, a > b, one solution. 

Trigonometrically, by our formulas, we would arrive at 
these facts, but a student who is not able to observe the 
geometrical relationships is not likely to be able to interpret 
the trigonometric formulas. When the sine of an angle is 
given, the angle may be either in I or II, a or 180° — a if a is 
either angle which satisfies the relationship. Then, 

sin 3 sin a • b sin a 

" = , gives sm /3 = ; * 

b a a 

if a < b sin a, sin /? will be greater than 1 and there is no 
angle satisfying the relationship ; if a > b, a > (3 (greater 
angle, greater side opposite), and only the acute angle ft can 
be taken ; if a < b, both values of ft can be taken. 

PROBLEMS 

1. Given a = 30°, a = 150, b = 60, 70, 75, 100, 150, 180, and 
200 respectively ; draw the figures and determine the number 
of solutions in each case. Solve for ft in each case where it 
is possible. 

2. Given a = 90°, a = 150, b = 75, 100, 150, 200. Discuss. 

3. Given a = 150, b = 75 ; a = 20°, 30°, 45°, 60°, 80°, 90°, 
120°, 150°. Discuss the solutions, geometrically and trigono- 
metrically. 



SOLUTION OF TRIANGLES 271 

4. Solve the following eight problems, having one or two 
solutions, and time yourself. Use the following form of solu- 
tion. The eight problems should be completed within one 
hour. 

a. Given a = 366, b = 677, a = 15° 10' ; solve for #, 

. Q b • sin a 

sin 6 = 

a 

log b = 
+ log sin a = 

— log a = 
log sin (3 = 

■ pi = 

or simply (3 = , if there is only one solution. 

b. Given a = 423, c = 288, y = 35° 15' ; find a. 

c. Given b = 376, c = 804, y = 68° 20' ; find p. 

d. Given b = 650, a = 830, a = 98° 56' ; find /J. 

e. Given a -67.2, c = 40.4, y = 24° 49' ; find a. 
/. Given 6 = .0188, c = .0196, y = 100° 14' ; find (3. 
g. Given a = 504.2, c = 1763, a = 12° 39' ; find y. 

h. Given 6 = 3,245,000, c = 2,488,000, (3 = 80° 28' ; find y. 

4. Type form : aba type with two solutions. — 

Form of solution when two solutions are found. 



Given a = 187, 6 = 235, a = 37° 15'. 

8in|8 = 6sinft ; c = aHiny ; check, c = & sin ?■ 
a sin a sin /3 

Or Newton's check formula, 

b + a _ cos | (ft — «) 

c sin ^ 7 

log 6 = 2.3711 
+ log sin « = 9.7820—10 
12.1531 - 10 
- log a = 2.2718 
log sin/3=" 9.8813- 10 



272 



UNIFIED MATHEMATICS 



ft = 49° 33' 




j8 2 = 130° 27' 




a = 37° 15' 




a = 37° 15' 




7i = 93° 12' 




72 = 12° 18' 




log a = 2.2718 




loga = 2.2718 




+ log sin 7! = 9.9993 - 


10 
10 


+ log sin 72= 9.3284- 


10 


12.2711 - 


11.6002- 


10 


— log sin a = 9.7820- 


10 


— log sin a = 9.7820- 


10 


log ci = 2.4891 




logc 2 = 1.8182 




ci= 308.4 




c 2 = 65.8 




log 6= 2.3711 




log 6= 2.3711 




+ log sin 7!= 9.9993- 


10 


+ log sin 72= 9.3284- 
11.6995- 


10 


12.3704- 


10 


10 


- log sin ft = 9.8813- 


10 


- log sin /3 2 = 9.8813 - 


10 


logCi= 2.4891 




log c 2 = 1.8182 





Compare with values for log c\ (and log c 2 ) found above. 

2 Ai = ab sin y 1 2 A 2 = ab sin 72 

log a = 2.2718 log a = 2.2718 

+ log b = 2.3711 log b = 2.3711 

+ log sin 7! = 9.9993 - 10 log sin 72 = 9.3284 - 10 

log 2A l = 4.6422 log 2 A 2 = 3.9713 

2^ =-43870 2^2 = 9360 

Ai = 21935 A 2 = 4680 

5. Given two sides and the included angle : type aby. — The 

method of solution here given is the solution by right triangles, 
since that involves no new formula and no greater amount of 
computation than the common solution employing a new tan- 
gent formula ; the tangent formula is given in Section 8 of the 

preceding chapter. 

Solution by right triangles. 




Given a = 280, 
b = 240, 



7 = 35°, 
a = 



tan a = 



a sin 35° 
— a cos 35 c 



a sin 7 
sin a 



check, c 



h = a sin 35°, 
x = a cos 35°, 
z — b —a cos 35°. 
b sin 7 



sin p 



SOLUTION OF TRIANGLES 



273 



2 A 

log a 

+ log sin 35° 

log h 

— log z 

log tan a 

a 

?■■ 
log a 

+ log sin 7 : 

— log sin a ■. 

logc : 

c : 

lOg & : 

+ log a ■- 
+ log sin 7 : 

lOg 2 J. : 
2 A: 

A 



ab sin y 
2.4472 
9.7586- 10 



4.5860 
38550 
19270 



log a = 2.4472 
+ log cos 35° = 9.9134- 10 
log x = 2.3606 



: 2.2058 


x = 229 A 


= 1.0253 


z = 10.6 


: 1.1805 




= 86° 13' 




: 58° 47' 




r 2.4472 




: 9.7586- 10 




2.2058 - 10 
■ 9.9990 - 10 
: 2.2068 

161.0 

2.3802 

2.4472 


Check. log b = 2.3802 

+ log sin 7 = 9.7586 

12.1388 

-logsin/3= 9.9321 

logc= 2.2067 


■9.7586-10 





- 



m 



m 



This problem also occurs frequently in surveying. It 
permits the determination of the direction and length of a 
tunnel through a mountain by means of the location of some 
point from which both ends of the 
proposed tunnel are visible. The dis- 
tance and direction from a given point 
to a second point, past some barrier, 
are determined by this method. Thus 
the distance from B to C through 
woods can be found, if some point can 
be located from which both C and B 
are visible. The distance and direc- 
tion to invisible points are constantly 
needed in artillery fire ; another method of finding distance 
and direction of the target is that of finding an observation 
point from which both the gun and the target, invisible at 
the gun, are visible to an observer. 



Distance across a barrier 



274 UNIFIED MATHEMATICS 

PROBLEMS 

Using the form of solution above, solve for the side opposite the given 
angle. 

1. Given a = 3846, b = 4977, y = 38° 10'. Find c. 

2. Given b = 4.832, c - 8.973, a = 108° 56'. Find a. 

3. Given a = .0485, c = .0682, £ = 58° 38'. Find h. 

4. Using the form of solution given above, find the side 
opposite the given angle in the following five problems ; time 
yourself, and compare with the time for the same five problems 
solved by the cosine law (page 255). 

a. Given a = 366, 6 = 677, 7 = 15° 10'. 

6. Given a = 423, c = 288, £ = 35° 15'. 

c. Given b = 627, c = 816, a = 100° 41'. 

d. Given a = 635, c = 341, (3 = 67° 38'. 

e. Given c = 184, b = 295, a = 130° 54'. 

6. Discussion of checks and methods. — The procedure by 
logarithms as with physical measurements involves numerous 
approximations. That two values of log c, in the check and in 
the solution, or by two different methods of solution, do not 
agree precisely is a frequent result of correct computation. 
However, the disagreement will be within certain well-defined 
limits, depending entirely upon the range (number of places) 
of the logarithm tables which are used both for numbers and 
for angles ; the tabular difference should be noted, mentally, 
and any discrepancy between check and computation should 
be examined as to its effect upon the value of the computed 
quantity. Thus no error in our computation (page 273) accounts 
for the difference between log c = 2.2068, and log c = 2.2067, 
nor does this here affect the value of c, However the 
angle of 86° 13' is so near to 90° that the tangent grows very 
rapidly ; the tabular difference here is large, and might easily 
affect our result, through the inevitable inaccuracy of ordinary 
interpolation in this neighborhood. 



SOLUTION OF TRIANGLES 275 

7. To determine the angles of a triangle when the three sides 
are given : abc type. — 



v 



tan - a = — ^~ 
2 s — a 

tan ! = -£-_ o - « + & + 

2 s - b 2 



(8 — a) (s — 6) (s — c) 



tan - 7 = A = V s(s — a) (s — 6) (s — c) = r • s 

2 s ~ c Check. a = 



= 

7 = 

a + +Y=" 

log (s - a) = 

+ log (s -b) = 

+ log (&• - c) = 

- log s =" 

log r 2 — 



a = 

= 

c = 

2s = 

s = 

s — a = 

s-b = 

s — c = 

Check by noting sum of s — a, w — 

— 6, s — c which equals s. 

log r == log r = 

-\og(s-a)=_ -log(s-6) = 

log tan -| a = log tan \ j8 = 

i«= 1/3 = 

a= = 

log r = log r = 

- log (s-c)= logs = 

log tan ±7 = logJ.=- 

17 = 
7 = 



.1 



Complete the solution of a problem of this type, form as 
above, taking a = 4320, b = 6840, and c = 8630. 

TIMING EXERCISES 

1. Employing the form of solution as above, solve the follow- 
ing four problems for the angles a, fi, and y, timing yourself ; 
write down the complete form necessary for the solution of 
each problem before using the logarithm table. 

a. Given a = 320, b = 640, c = 580. 

b. Given a = 44.8, b = 76.2, c = 70.4. 

c. Given a =4.49, 6 = 8.87, c = 9.13. 

d. Given a = .0624, b = .0688, c = .0731. 



276 



UNIFIED MATHEMATICS 



2. Solve the following four problems for a, /3, and y, taking 
note of the time required. 

b = 640.6, c = 580.4. 

b = 7461, c = 5395. 

6 = 2.346, c = 3.045. 

b = 2.34 x 10- 6 , c = 2.87 x 10~ 6 . 

3. How would the solution of problem 2 cl be changed if a, 
b, and c were given as 1.43, 2.34, and 2.87, respectively ? 



a. Given a = 320.4, 

6. Given a = 3482, 

c. Given a = 1.835, 

d. Given a = 1.43 x 10~ 6 , 



PROBLEMS 

Type problems. Solve for the other parts 
required for the solution of each problem. 

b 

6788 
6788 
6788 
6788 



make a note of the time 



1. 
2. 
3. 

4. 

5. 

6. 

7. 

8. 

9. 
10. 
11. 



a 
8294 
8294 
8294 
8294 
8206 
8206 
8206 

.03267 



a 
33° 15' 



33° 15' 



6009 
356 



9645 



235 
.05431 



33° 15' 



67 c 
33 c 



25' 
15' 



133 c 
64° 



15' 
10' 



33° 15' 



67° 25' 



63° 40' 



83 X 10 6 , 67 x 10 6 , 54 x 10 6 . 

Given two forces of magnitude 384 and 276 acting at an 
angle of 38° with each other. Find the angle which the re- 
sultant makes with the larger force and the magnitude of 

the resultant. Note that the 
problem is simplified by re- 
garding the line of the larger 
force as an axis of reference ; 
the second force adds to this 
a component 276 cos 38° ; the 
276 sin 38° 




Resultant of two forces 
vertical component is 276 sin 38 



tan <£ = 



384 + 276 cos 38°' 
the magnitude of the resultant, r, is by the sine law, 



SOLUTION OF TRIANGLES 277 



276 



sin (180° -38°) sin<£' 

, 276 sin 38° 

whence r = : ; 

sm <f> 

or r = V(276 sin 38°) 2 +(384 + 276 cos 38°) 2 , 

as hypotenuse of a right triangle ; or 

r = V(276) 2 + (384) 2 - 2 x 384 x 276 x cos 38°, 

by the cosine law. 

12. Given two forces of magnitude 684 and 450, acting at an 
angle of 64°. Find the resultant in magnitude and direction, 
graphically and trigonometrically. 

13. From an aeroplane moving horizontally at rate of 120 
miles per hour (176 feet per second) a bullet is shot at right 
angles to the path of the aeroplane with a velocity of 2800 
feet per second. What is the resultant velocity in magnitude 
and direction ? 

14. A cylindrical trough of horizontal length 20 feet and 
with the ends semicircles of radius 2 feet each, contains how 
many gallons of water for 1 foot in depth, for 1J feet, for 2 
feet? 

15. A typical oil tank is 30 feet long and has a diameter of 
8 feet. Compute the volume in barrels (see page 94) for each 
foot of depth. Do not carry beyond tenths of a barrel. 

16. What angle does y = 2 x + 12 make with the se-axis ? 
What angle does 3 y — 4 x — 20 = make with the a>axis ? 
Find the area and the other angle of this triangle, formed by 
the two lines and the #-axis, by trigonometrical methods. 
Find the angle between the two lines by the formula, 

tan <j> = m * ~ ?? 2 and check. Find the area by analytical 
1 + mjn.2 

methods to check upon the trigonometrical solution. 



278 



UNIFIED MATHEMATICS 




Moving aero- 
plane 



17. In the figure AB represents the move- 
ment in 30 seconds of an aeroplane moving 
parallel to OX at rate of 120 miles per hour ; 
Z A OX is measured as 83° 30' and Z BOX is 
measured as 74° 48' ; find the distance OB 
and the position of the aeroplane at the end 
of the next 30 seconds if it continues on its 
present course. 

18. In problem 17 discuss the percentage 
effect of an error of 1° in angle AOX and in 
angle AOB, respectively. 

19. Given that observers A and B at the ends 
of a battleship 340 feet long observe an object 
O at angles of 106° 30' and 72° 48' respectively 
with the line AB. Find BO and AO. Solve 
this problem also graphically. 

20. A and B are observation stations on the shore, 10 miles 
apart and may be assumed to be on an east and west line ; R, 
the battery, is three miles east 

of A and one mile south. Given 
that a battleship P is observed 
from A at an angle of 68° 12' 
and, from B at an angle of 
59° 02' with AB ; find the coor- 
dinates of P; find the range 
from R and the angle made by 
RP with the east and west line ; 
find the distance using the for- 
mula for the distance between 
two points. This problem is 
solved in actual practice graph- 
ically on large plotting boards. Using -j- inch to the mile, how 
closely could you approximate the distance ? 

21. In the preceding problem, suppose that the observa- 
tions reported at the end of 1 minute are from A, 68° 08', and 



S 



2:i3: 



~M 



10. 



Shore battery observations 



SOLUTION OF TRIANGLES 



279 



from B, 59° 12', locate the direction of movement of the 
ship. 

22. If the battleship of the two preceding problems is 
600 feet long and broadside to A, what angle does it subtend 
at A? 

23. When guns are tested at Sandy Hook or other proving 
grounds, the actual range for any angle of elevation of the 
gun is obtained by coincident observation of ' d splash of 
the shell by several observers located in toweics along a line 
roughly parallel to the line of fire of the projectile. The 





^' 1 


i 


' — 


1 1 




pi n 




















- ^V^ — 












=^~ 


x^p= 


i 


Lin 


































h\ 1 i 














\\ ! i ' 














6 ~\ ' 














^\ 




































■ ! T-pV 1 ' 
























i \\ p> 1 r 




D 




E 






\\ i : 
















<-4~40.C 


i 






— M- 


i 


-p; 















! ' 






Ml 


i 






III 


i 




. ,,. , 






i 






! ! 




1 1 



Observation towers at proving grounds 



shell is loaded with a slight charge of high explosive in order 
to give a splash of some magnitude. For convenience in our 
figure, we have assumed the towers on a north and south line 
spaced as indicated ; the height of the tower is regarded as 
negligible. Given that observers at A, B, C, D, and E observe 
the angle of the splash with the north and south line, the 
azimuth angle, as 2° 20', 21° 24', 27° 16', 41° 35', and 68° 54', 
find the distance and direction of S, the splash, from G, the 
gun, which is 400 yards due east of A. Xote that any two 
observers give the position of S ; the substantial agreement 
of three observers is taken as sufficient. Compare the solu- 
tions. Determine the coordinates of S with respect to a 
horizontal axis through BCDE and a vertical axis through 
AG. Assuming that the gun was pointed south the deflection, 



280 UNIFIED MATHEMATICS 

to the east here, is termed the drift ; it is due to wind and 
other causes ; determine this angle. 

It is not essential that the towers be in a straight line ; the 
distance and direction of lines between adjacent towers is 
carefully measured. A large plotting board is used to obtain 
a graphical solution. 

24. Employing the general form of solution, adapted to each 
case, solve the following triangles for the unknown sides or 
angles, mak.. v note of the time required : 

a. Given a = 44.82, b = 76.24, c = 70.48. 

b. Given a = 366.5, b = 677.9, y = 15° 11'. 

c. Given b = 376.3, c = 804.8, y = 68° 27'. 

d. Given b = 816.4, p = 67° 17', y = 34° 9'. 

e. Given a = 915.5, y = 90°, a = 32° 3'. 




London Bridge 

Five pure elliptical arches ; central one 152 feet by 37 feet 10 inches, 
and the others 140 feet by 37 feet 2 inches. 



CHAPTER XVIII 



THE ELLIPSE 

1. Parametric equations of an ellipse. — The parametric 
equations of the circle, with center at the origin 

x = r cos 0, 
y = r sin 6, 
and of the circle with center at (h, Jc) 
x — h = r cos 0, 



if modified to read, 



and 



y — k = r sin 0, 

x = )\ cos $, 
y = r 2 sin 0, 

x — h = i\ cos 0, 
y — k = r 2 sin 0, 



respectively, give a curve which is closely related to the circle. 
This curve is called the ellipse ; r l and r 2 are called the major 
and minor semi-axes of the ellipse, and the circles obtained 

281 



282 



UNIFIED MATHEMATICS 



with radii r x and r 2 are called the major and minor auxiliary 
circles of the ellipse, assuming r x greater than r 2 . 




An ellipse with its major and minor auxiliary circles 



x = ri cos 6 
y = ri sin 6 



gives the larger circle. 
x = r x cos 6 



x = r 2 cos 
y = r 2 sin 



gives the smaller circle. 



gives the ellipse. 
y = r 2 sin 

P, Q, and B are called corresponding points. 



Eliminating 6 between the two parametric equations of the 



ellipse gives 



— -f iL - co S 2 + sin 2 = 1. 



Writing, as is customary, a and b for r x and r 2 this equation 
becomes 

a 2 6 2 ~ 



THE ELLIPSE 283 

If the ellipse had the center (Ii, k) and the two values a and b 
corresponding to r x and r 2 , the equation of the ellipse would 
be written 

(x — 7i)= a cos 6, 

(y — k) = b sin 0, 
in parametric form ; and 

(x-hy (y-fc) « =1 
a 2 6 2 

in so-called standard form. If we assume, as we have above, 
that a is the radius of the larger circle, we would have for an 
ellipse placed vertically, 

x — h = b cos 0, 

y — k = a sin 0, 

(x-h Y (y-k) 1== 
6 2 a 2 

as the equation of an ellipse whose major axis is vertical. 
Xote particularly that the terminal side of the angle does 
not pass, in general, through the point on the ellipse which 
corresponds to that angle, but the angle is made by a line 
passing through the corresponding points on the major and 
minor auxiliary circles. 

PROBLEMS 

1. Construct the ellipse 

x =10 cos 6, 

y = 6 sin 6, 
by finding the points on the ellipse given by = 0°, 30°, 45°, 
60°, 90°, and the corresponding points in the other quadrants. 

2. Construct the ellipse 

x = 10 cos 0, 

y = 6 sin 6, 
using corresponding points on the major and minor auxiliary 
circles to locate points on the ellipse. 



284 UNIFIED MATHEMATICS 

3. How does the ellipse 

x— 3 = 10 cos 0, 
y + 2 = 6sm6 7 
differ from the preceding ellipse ? 

4. Locate 10 points on the ellipse 

x — 3 =6 cos $, 

y + 2 = 10 sin $, 
and draw the curve, 

5. Prove that a is the largest value and b the smallest of a 
line from to any point on the ellipse. This gives the reason 
for the names, major and minor axes. 

6. Show that every chord of the ellipse through is bi- 
sected ; is the center of the curve. 

7. The five arches of the London Bridge are of true elliptical 
shape; the central arch has the width 152 feet (=2 a) and 
the height 37 feet 10 inches (=6). Find the equation and 
plot 12 points on this arch ; b may be taken as 38 feet. The 
adjacent arches are of length 140 feet and height 37 feet 2 
inches. Write the equations. In each case take the major 
axis as #-axis and minor axis as 2/-axis. 

2. Properties of the ellipse. — The equations 
x = a cos 6, 
y = b sin 6, 

wherein a>b, can represent any ellipse whatever, when the 
axes of the ellipse are taken as the axes of reference. The 
geometrical peculiarities of this ellipse will be characteristic 
of all ellipses, provided, of course, that no limitation is placed 
upon a and b (except that a may be taken as greater than b). 
Each ordinate of the ellipse 

x = a cos 0, 
y = b sin 0, 



THE ELLIPSE 



285 



is a constant proportional part of the corresponding ordinate 
in the major auxiliary circle, 

x = a cos 0, 
y — a sin 0. 











_^_ r5-- :S .5+- :: — 




it-'Xi** " s >. 


VV ^N, 






ijj"!2 in 






f*2 ' >7? 




:- _ a±- - MQ . ( ^^y-rP?(»?4a ::::^:X 








A?w , %^ ' >^ -n\y' ""- \'f\ 


r^x '''1/ ^'v rT x i 




-/-^ "A-M - --- S - L \ ' 






. ::::::b:::::::2_::-s^::j:::::E::::-:::::::!S.:: 




fv : ■C ^ ^N^- F r -5 






IVr Vlr^ V^V6> / i^/l 


ipfcr^n ^*(r* \- ^ N 1 ' v f v 


VrK/j^, j^7a[<? \\s x Ah , *Mt / 


V rsK >- Vx]^ v^k 1 > 't y<>S>>' 7 




\ L' S K'^ ^iV^rt . Jilfty^ 'f,^ X / 


-----^^---"^W-fekiw^ft" 4-1.— 


------4 = = = = BB ! > = --S_. ^ *__ 


_#____i? :: :::::^:::^:::_::::: 


:::::::±: s ? JV 


s r> -Y 


\xK aI 


^k&£ ::: :::: :::::;z:__ __:__:::::: 


:::: s <r^r \^"^- 


■^L^/rt ACi a ." l1i ^< 


^4iVos'^--?/-^>^>--'^ - - 




. .. . .. _L_ _^ 




It 


± ± - 



An ellipse with its major and minor auxiliary circles 



X = r x cos 6 
y = ri sin 6 



gives the larger circle. 
X = Vi cos 



x = r 2 cos 6 
y = r 2 sin 6 



gives the smaller circle. 



gives the ellipse. 

2/ = r 2 sin(/° 

P, Q, and .R are called corresponding points. 



Take any point (a cos 9, b sin 6) on the ellipse, the 
corresponding ordinate in the circle is y c = asin9; but 

y e =b sin = - (a sin 0) = - ■ y c . Conversely, if the ordinate of 

any point on a given curve is always a constant proportional 



286 



UNIFIED MATHEMATICS 



part of the ordinate of a corresponding point on a given 
circle, for equal values of the abscissa, the curve is an ellipse ; 
the ratio need not be less than unity as the figure clearly 
indicates, a vertical ellipse being represented when the ratio 
is greater than unity. This property of this ellipse is equally 
true of any ellipse, replacing the terms ordinate and abscissa, 

where given or implied, 
by perpendiculars to the 
major and minor axes, re- 
spectively, of the curve. 

If the major auxiliary 
circle is rotated about 
OX as an axis through 

an angle a, cos a = - , the 
a 

projection upon the plane 

of the original position 

will be the ellipse 

x = a cos 6, 
y = b sin ; 

evidently the x of any 
point on the projected 
curve may be taken as the 
x of a point on the project- 
ing circle, or x = a cos 6. 
The ordinate on the pro- 
jected curve is in a con- 
stant ratio, cos a — - , to 
jj a 

Elliptical section of a circular cylinder t j ie ordinate on the circle. 

Further any plane section of a cylinder with circular base is 
an ellipse, for with the same abscissas, the ordinates of the 

curve of section bear a constant ratio - — to the correspond- 




cos a 



ing ordinates on the circular base. 



THE ELLIPSE 287 

Let the plane cut the axis of the cylinder in 0, and let 
BOX, on the diagram, be the intersection of the cutting 
plane with the plane of the circular section through (paral- 
lel to the base of the cylinder). Note that angle PMQ is 
constant. 

Similarly if a circular cone is cut by a plane cutting all the 
elements of the cone, the section formed is an ellipse; the 
geometrical proof is rather too complicated to give here but an 
analytical proof is indicated in Chapter XXXII, Section 6. 
The ancient Greeks studied the properties of the ellipse entirely 
from the point of view of the curve as a plane section of a 
cone. The Greek theoretical work concerning the properties 
of conic sections made it possible for Kepler to discover that 
the path of the earth about the sun is an ellipse, and for 
Newton to formulate the law of gravitation. 

The properties mentioned are intimately connected with the 
applications of the ellipse in engineering problems. 

From the property of the ordinates it follows that the area 

of an ellipse is - • ira 2 = -n-ab. A = irab. 
a 

The proof is strictly by a " limit process," and may be made 
reasonably evident by dividing the semi-axis into 25, 50, 100, ••• 
equal parts and drawing two series of rectangles on these 
equal subdivisions about the corresponding ordinates to fall 
entirely within (or entirely without) the ellipse and the circle, 
respectively. As the subdivisions are increased in number the 
one series of rectangles has the area of the quarter-ellipse as a 
limit ; it differs from this area never by an area as great as the 
rectangle of height b and base one subdivision ; so also the 
sum of the second series of rectangles has the quarter-circle 
as a limit, never differing from it by an area as great as the 
rectangle of height a and base one subdivision ; but the sum 

of the series of smaller rectangles always equals - times the 

a 

sum of the series of larger rectangles since the bases are equal 



288 UNIFIED MATHEMATICS 

and the altitude of any one of the smaller is - times the alti- 
tude of the corresponding one of the larger ; evidently, then, 
the limits of these sums are in the same ratio. In the dia- 
gram above the ruling of the paper divides the semi-major 
axis into 25 parts, and gives the rectangles. 




The Colosseum in Rome 

A famous elliptical structure, 615 feet by 510 feet, by 159 feet high ; 
the arena is an ellipse 281 by 177 feet. 

A rectangle of dimensions 2 a and 2 b may be circumscribed 
about the ellipse. The sides of this rectangle are tangents at 
the vertices of the ellipse ; the middle lines parallel to the 
sides are the axes of the ellipse ; the horizontal sides of this 
rectangle cut the major auxiliary circle in points which cor- 
respond to four symmetrically located points on the ellipse ; 



sin = ± -, cos = ±\/l - - = ± — -, whence 

a * a 2 a 



ir- 



x= ± Va 2 — b 2 , y = ± — , are the coordinates of these points on 

a 

this ellipse. The points on the major axes ( ± Va 2 — b 2 , 0) 
are the foci of the ellipse, and enjoy special properties with 



THE ELLIPSE 



289 



respect to points on the ellipse, 
the path of the earth. 



The sun is at one focus of 



3. Standard definition of the ellipse. — The locus of a point 
which moves so that its distance from a fixed, point called the 
focus is in a constant ratio, e, less than 1, to its distance from 
a fixed line called the 
directrix, is an ellipse. 

Let the fixed point be 
i^and D'D the fixed line. 
Through F drop a per- 
pendicular to the direc- 
trix D'D meeting it in 
R and use this line as 
By definition, 





Jy. 








ft i = = = = = == = »»*- ^^Hs 










-T 4 £ L 


-T ' /t \ 


D± [£• 





A A' 

PF = e-PZ defines the ellipse 

The constant e is called the eccentricity. 



#-axis. 

then, the ellipse will be the locus of a point which moves so 
that PF = e • PZ, wherein PZ is the perpendicular from P to 
the directrix. Two points of our curve will be found to lie 
upon the a>axis ; the two points are the points A and A' which 
divide the segment FR internally and externally in the ratio e 
(taken asf on our figure). Take the middle point of A' A as 
the origin, designating A' A, which is a fixed length dependent 
upon FR and e, by 2 a. Then OA = OA' = a. 

AF=e-AR. 
A'F=e-A'R. 

AF+ A'F=e(AR + A'R). 

2 a = e(AR + OA' + OR). 

= e(OA + AR+OR) 
= 2 e • OR, whence 

OR = -- 

e 

A'F- AF= e(A'R - AR). 
2 OF=2ae. 
OF=ae. 

F is (_ ae, 0) and D'D is x + - = ; 

e 



290 UNIFIED MATHEMATICS 

PF = e • PZ, taking P as (x, y), gives in analytical language, or 
formulas, 

V(x + aef + {y - 0)2 = e ■ /» + "! = ea 4- a. 



a£ + a 2 6 2 _|_ ^2 _ ^.2 _j_ a 2 t 
aj2(l_ e 2) +2/ 2 = a 2(l_ e 2) < 



# 2 ?/ 2 



| 2 = 1. 

a 2 a 2 (l - e 2 ) 
Let b 2 = a 2 (l - e 2 ), 



whence — I- sL = i. CI) 

a 2 ft 2 V ; 

But this equation is satisfied by every point 

a; = a cos 0, 

2/ = b sin 0, 

determined as we have indicated above, and conversely ; our 
two definitions are equivalent to each other. 

The form of equation (1) shows that the curve is symmetrical 
with respect to the two lines which we have chosen as axes ; 
i.e. since x and — x give precisely the same equation to deter- 
mine y, the curve is symmetrical with respect to the y-axis ; 
similarly since y and — y give precisely the same values for x, 
the curve is symmetrical with respect to the a>axis. The 
coordinate axes are axes of symmetry of the curve. 

By symmetry with respect to the y-axis we mean that if the 
right half of the curve is folded over on the y-axis as an axis, 
i.e. revolved about it as an axis (or axle) through an angle 
of 180°, the two sides will coincide throughout. Hence 
corresponding to F x and D'D, there is another focus F 2 and a 
corresponding directrix D 2 D' 2 , enjoying precisely the same 
properties with regard to the curve as F 1 and D'D, 



By symmetry, 
and since 

Similarly, 
and, since 



THE ELLIPSE 
P.Fy = P 2 F,, P 1 Z 1 = P 2 Z 2 , 
P& = e • P X Z X , P 2 F 2 = e • P 2 Z 2 . 
P.P, = P,F 2 , P 2 Z, = P,Z 2 , 
P 2 F X = e • P 2 Z 1} P X F 2 = e • P X Z 2 . 



291 











■ ■ ■ : 


nil 


rrr^r- 


44+4- 

















JJz— 














■ 




— ! — 1 — j — ' 










1 




























1 






















, 






^2~— 






*pbl=F 


'-* 




^— F ' ' ; 




-^T^Zr 




■ 






— =^ 




- — - — 1 








""■V. 














V^ ' 










^v 








- 




-y 


•^ ; 




^N 


*<£?- 





















—f 
















\g ' 













/ 




























-f— 


7n/ 


-,^r e 


p^ILL 






v *^ t ^ 




\^ 










"1 


f— 




(V>^ 






-H — 




^^«<^ 


J — Yi. 
























































































-£"?■ 


Ai 


<<-, 








O 






1 1 :j<. 


• 'r 


:~K-, 






















j 




























/ 




























/ 




























/ 




























/ 




























/ 




























/ 






















































s^ 














































































































































--/)■' 




j"j' 2 












































1 1 1 1 


i 


Mil 


MM 




1 1 1 1 





Symmetry of the ellipse with respect to its axes 

4. Sum of the focal distances constant. — Taking any point 
Pife 2/i) 011 the ellipse, the focal distances P X F X and P X F 2 are 

equal to a — ex x and a + ea^, respectively. 

For PiFi = e-PxZ = elx 1 ) = ex 1 — a, which is a negative 

distance since P x and are upon the same side of the line 



D'D. Similarly P X F 2 = e ■ P X Z 2 = e 



»i + 



-1 



— (exi + a). As 



positive values PiF x and P\F 2 are a — ex 1} and a + e^ ; these 
may also be derived by the formula for the distance between 
two points, noting that (x ly yi) is on the curve ; 



PFi + PFo = a — exi + a + ex x — 2 a. 



292 



UNIFIED MATHEMATICS 



5. Right focal chord. — When 



x — ae, 



aV* 2/2_ 1 



62 



t 



¥ 



b\l - e 2 ) = - 9 , since b 2 = a 2 (l - e 2 ), 



b 2 2 b 2 

y = ± — , giving — ■ as right focal chord. 
a a 

The value y = — is the value of y obtained in the parametric 

form when sin = - , ?/ = & sin = — • 
a a 

x = a cos 6 = ± a\/l — — = ± Va^^P — ± ae 
* a 2 
showing that the focus as we have now denned it coincides 
with the focus as first defined. 

6. Standard forms of the ellipse. — The equation ~ + ^ = 1 
may be interpreted geometrically, 

~CM 2 MP 2 = 1 
CA 2 CB 2 ' 



a 2 b 2 




in which CM and MP represent the distances cut off from the 
center on the major and minor axes of the ellipse by the per- 



THE ELLIPSE 293 

pendiculars to the axes from any point on the ellipse ; CA and 
CB are the lengths of the semi-major and semi-minor axes. 

Given a horizontal ellipse with center at (h, k) and axes a 
and b, the relation 

CM 2 MP 2 = 1 

04 2 CT 2 

becomes 

^ *- + ^ £■ = 1, since 

«2 T 62 

(73/ = a? — h and 3/P = y-k. 

Similarly a vertical ellipse with center (h, k) and axes a and 
b, respectively, has the equation 

a 2 &2 

(73/ is here y — k and MP is a; — h. 

Type .forms :^ + (rf = l; k^ + fczffi.L 

J± ^ a 2 & 2 « 2 6 2 

& _ a 2(l _ g 2) . 

. [horizontal . .. \y — k = 

loci ot <( ellipse on the line { ^ 

i vertical i x — fi ^= u 

at a distance ae = Va 2 — 6 2 , from the center (7i, A:) ; 

right focal chords extending a distance — { * 

a | horizontally 

from the foci on either side of the major axis I ^ ~ 



7. Limiting forms of the ellipse equations. — 
* — ~ ' + ^ ~~ ' = 1 represents an ellipse. 

' — ~ ' + ^-^ — *- = represents a point ellipse or two 

imaginary straight lines through (h, k) ; the only real point 
which satisfies this equation is the point (h, k). 



294 UNIFIED MATHEMATICS 

( x — k) 2 (y — k) 2 

*- '- ■+- — = — 1 represents an imaginary ellipse ; 

only imaginary values of x and y can make the sum of the two 
squares on the left equal to — 1. 

The equation of any ellipse with axes parallel to the coordi- 
nate axes may be written, 

( X - X V + fo-^ 2 = k, wherein 
a 2 b 2 

ka 2 and kb 2 represent the squares of the semi-axes. As k 
approaches zero the ellipse diminishes in size, and when k = 
the equation represents the point (x 1} y y ) ; when k becomes 
negative the ellipse becomes imaginary. 

8 . Illustrative problems. — 

a. Plot the ellipse 4 x 2 + 16 x + 9 y 2 - 18 y - 75 = 0. 
First write in form to complete the squares, 

4(x2 + 4 x ) + 9(j/ 2 -2y ) = 75. 
Complete squares : 

4(x 2 + 4x + 4)+ 9(?/ 2 -2?/+ 1) =75 + 16 + 9. 
4(x + 2)2 + 9(> - l) 2 = 100. - 

Write in standard form : 

(x + 2)2 { (y -!)» _., 



25 H* _ 

9 3 



(-2,l); a2 = 25;^ = f ;a e=V25-f=V 1 f- 



62 _ lop _ 20 
a 45 9 ' 
Plot the center, extremities of major and minor axes ; foci ; extremities 
of right focal chords ; at least one further point, obtained from the 
original equation and selected so as to give a point approximately midway 
between the extremity of a focal chord and the corresponding extremity 
of the minor axis ; by symmetry three other points are obtained. In this 
case x = gives desirable further points. 

9 + V81 + 675 

V 9 

= 1 ± iV756 

= 1 ± K27-5) 

= 1 ± 3.06 

= 4.06 or - 2.06. 



THE ELLIPSE 



295 



b. Plot the elliptical arch of a bridge, arch 125 feet wide and 
37 feet high. Plot points for every 10 feet of the span ; and 
compute to -^ of 1 foot. 




Elliptical arch, 125 foot span by 37 feet high 

Scale, 1 inch to 40 feet. Horizontal measurements are from the middle 
of a quarter-inch square. 



,/•- 



(62.5) 2 
62 = 372 



372 



372 

62.5' 



a2 = (62.5)2 ; 62 = 372 . ae = v/(62.5)2_372; 

372^1-^-V 
62.52/ 

f^y.,2. 

\62.5/ 
and multiply the latter by x 2 = 100, 



?/ 2 = 37 2 

Here compute 37 2 and ( -^-Y 
V62.5; 

400, 900, 1600, 2500, and 3600 ; extract the square root of the difference ; 
use four-place logarithms. 

log 37 = 1.5682 
log 62.5 = 1-7959 
log quot. = 9.7723 — 10 
log quot. 2 = 9.5446 — 10 



/37 \ 2 _ 

(62.5 ) = 



3504 



yw 2 = 


1369 - 


35.04 


2/io 


= 4-36.51 


2/20 2 = 


1369 - 


140.16 


2/20 


= +34.93 


2/30 2 = 


1369 - 


315.36 


2/30 


= +32.46 


2/40 2 = 


1369- 


560.64 


2/40 


= + 28.43 


yso 2 = 


1369- 


876.0 ; 


2/50 


= + 22.20 


y*o 2 = 


1369- 


1261.44 ; 


2/60 


= + 10.37 



296 UNIFIED MATHEMATICS 

Exercise. — Draw the major auxiliary circle on half-inch 
coordinate paper, and compute corresponding ordinates in the 

37 

preceding arch as of the ordinates on the circle ; e.g. for 

62.5 

tyrr 

x = 10, find y graphically on the circle and multiply by r^-r- 



PROBLEMS 

1. Plot the ellipse 9 a 2 + 36 a; + ?/ 2 — 6y = 0; what are the 
coordinates of the foci ? 

2. Plot one quarter of the ellipse — — + — = 1. 

H F 147 2 59 2 

3. Plot an elliptical arch, width 233 feet, height 73 feet, 
plotting at least 10 points spaced at 20-foot intervals from the 
center. These are the dimensions of the arch of the Walnut 
Lane Bridge in Philadelphia; the arch is approximately an 
ellipse. 

4. Plot the upper half of an ellipse, giving a vertical ellip- 
tical arch 13 feet li inches high, and 9 feet 2f inches wide. 
This represents the upper portion of an elliptical sewer used 
in the city of Philadelphia. 

5. What limitation is there upon the values of A and B if 

the equation, Ax 2 + By 1 + 2 Gx + 2 Fy + C = 0, is to represent 
an ellipse ? 

6. Put the following equations in standard form, complet- 
ing the square first and reducing to standard form by division. 
Time yourself. 

a. 4x 2 + 9y' i -$x + 36y = 0. 

b. 3 a 2 + 24 a + 2/ 2 -62/ -43= 0. 

c. 5x*-17x + 10y 2 -100 = 0. 

d. 5x 2 + 122/ 2 -117 = 0. 

e. 3^-24x + 42/ 2 + 16y-52=0. 

7. Plot the preceding five ellipses, choosing an appropriate 
scale. Plot the extremities of major and minor ;*xes, the 



THE ELLIPSE 297 

extremities of the right focal chords, and one other point 
obtained by computation, together with the three points sym- 
metrical to the computed point. 

8. Determine a 2 and b 2 to one decimal place, in the follow- 
ing three ellipses : 

a. 17 x 2 + 43 y 2 = 397- 

b. 5x 2 -17x + 10y 2 -35y = 0. 

c. 7(x - 2) 2 + 3(2/ - 3) 2 = 39. 

9. In the three ellipses immediately preceding determine 

ae, — , and - to one decimal place. 
a e 

10. In each ellipse of problem 8 determine x when y = 2. 

11. Using the data of problems 8, 9, and 10, plot the three 
ellipses of problem 8. 

12. In the ellipse £— + %- = 1, find the foci. What is the 

* 100 36 

distance of the point whose abscissa is -f- 3 from each focus ? 
of the point whose abscissa is 4, 5, 6, 7, x 1 ? 

13. Put the following equations in standard form and dis- 
cuss the curves : 

a. x 2 -6x + y 2 +8y-10=0. 

b. x 2 -6x + 42/ 2 +8y + ll =0. 

c. z 2 -6a + 4?/ 2 +8?/ + 13 = 0. 

d. x 2 - 6x +4# 2 + 8# + 14 = 0; 

e. x 2 -6x + 4:y 2 +8y + k = 0. 

14. The path of the earth about the sun is an ellipse with 
eccentricity .01677 ; this may be taken as -g- 1 ^- in the following 
computations. If the major axis of the earth's orbit is 185.8 
million miles, determine the focal distance, i.e. from the sun to 
the center of the path ; determine also the minor axis. If a 
scale of one-half inch to fifteen million miles is taken, at what 
distance will the point representing the sun be from the center 



298 



UNIFIED MATHEMATICS 



of the path ? What will be the difference in length between 
major and minor axes? 



^-4- ' " " " " 


_I_ 






















_S k 


s s 


JJA. v v 










rtn One juai ,erojea '/ft if\ 


ou i i lt j- U i I \ 


Q> OiC. \ 






,/i.K. „J[ i i i-U-LJ, ' ; o ij-ol. V 


4-o ce—rrnoi^; a— yztwt; c 


J mi/W'7i4)J JJ (iivfi- V 


fi — .Jin,^ n— t^2:M» ^ 




r; 7v- k- 








r 


-i-*- \ 


-W- 


_r _ 




A 3 J 






r5 ^3 r C JL5 zp,C 7:S: ..iJU ±0:5: ti. 


r i in i r iir i 




. .. MM f lit 



The path of the earth about the sun 

The distance of the sun from the center of the ellipse is represented 
by gV of an inch, on this diagram. 

9. Tangent to ellipse of given slope. — (See page 225.) To 
find the tangent of slope 2, y = 2 x + k is solved as simnltaneons 
with the ellipse equation. An equation whose roots are the 
abscissas of the two points of intersection is found and the con- 
dition is used that the two points of intersection be coincident. 






y = 2x + k, 
b 2 x 2 + a 2 (4 x 2 4- 4 kx 4- k 2 ) - a 2 b 2 = 0. 
x* (b 2 4- 4 a 2 ) + 4 a?kx 4- a 2 k 2 - aW = 0. 



2 a 2 fc ± V4 a 4 fc 2 - (fr 2 + 4 a 2 )(a 2 A; 2 - a 2 fr 2 ) 



6 2 4- 4 a 2 



2 a 2 fc ± Va 2 5 4 4- 4 a 4 ft 2 - a 2 b 2 k 2 
6 2 + 4a 2 



= - 2 a 2 fc ± Va 2 b 2 (b 2 4- 4 a 2 
6 2 4-4a 2 



A: 2 ) 



THE ELLIPSE 299 



Put b 2 + 4 o 2 - ft* = o. 

£2 = 52 + 4 a 2 ? 



fc = ± V& 2 + 4 a 2 , 



2/ = 2.r±yV + 4,7 2 
are the two tangents of slope 2. 

Similarly the tangent of slope m is obtained by solving 

(1) E? + ^ = l and 
K) a 2 62 

(2) y = ma; + ft 

as simultaneous, and writing the condition for equal roots. 
Clearing (1) of fractions, 

&2 a .2_|_ a 2^2_ a 2 & 2 = 0j 

substituting from (2), 

VW + a 2 ( m 2aj2 + 2 fcma + A; 2 ) — a 2 6 2 

= (6 2 + a 2 m 2 ) x 2 + 2 ka 2 mx + (a 2 & 2 - a 2 6 2 ) = 0. 
, _ - ka 2 m ± ^/k 2 a 4 m 2 —(b 2 + a 2 m 2 ) (a 2 k 2 - 

X — ■ ; 

b 2 + a 2 m 2 
Putting the discriminant equal to zero, 
k 2 = b 2 + a»ro 2 . 



ft = ± V6 2 + a 2 m 2 , whence 



y = ??i# ± -\/a 2 m 2 + 6 2 
are the two tangents of slope ??i to 

a 2 6 2 

This method of obtaining the tangent applies to any curve 
given by an equation of the second degree. 

10. Focal properties of the ellipse. — The perpendicular from 
the focus (ae, 0) on any tangent of slope m meets it in the 
point, whose coordinates are found by solving the equations of 
these lines as simultaneous. 



300 



UNIFIED MATHEMATICS 



The equations, y = mx + ^/a 2 m 2 + b 2 , of the tangent, 



and 

may be written 



y — = (x — ae), of the perpendicular, 



y — mx = V« 2 m 2 -J- b 2 
my + x— ae. 




The perpendicular from the focus upon any tangent to an ellipse meets 
it on the major auxiliary circle 

The point of intersection satisfies both these equations ; 
further it satisfies the equation obtained by squaring and add- 
ing both members of each of these equations : 

(1 + m 2 )y 2 + (1 -h m 2 )x 2 = a 2 m 2 + b 2 + a 2 e 2 

= a 2 m 2 + b 2 + a 2 — b 2 
= a\l + m 2 ). 
x 2 + y 2 = a 2 ; 

hence the point of intersection of the perpendicular from the 
focus on any tangent lies on the major auxiliary circle. 

Note that the above demonstration applies equally well to 
the perpendicular from the other focus (— ae, 0) and equally 
well to the other tangent of slope m, y = mx — s/ahn 2 -f b 2 . 



THE ELLIPSE 



301 



11. Tangent to an ellipse at a point Pi(x u y x ) on the ellipse. — 

The method outlined is general, being applicable to any alge- 
braic curve. The point P 1 (a5 1 , y{) on the curve, considered as 
rixed during the discussion, is 
joined to a neighboring point 
P 2 (x l i-7i, t/i + fc) on the curve, 
which second point is then made 
to approach (a^, y t ) along the curve. 
The slope of the chord joining P 1 

k 
to P 2 < —, is found not to change 
h 

indefinitely, but is found to ap- 
proach a definite limiting value as 
h and A: approach zero, 



i.e. as 



.---!-, . 


X 


y 




t 


::::::::::::::::#£::::::::: 


:::::^::::::,|J35zZ5SU: 


::::::tiSE^ ? :-J:::::::::::: 


::::::::::^f:f:i:::::::::::: 


1 j 


+ % 






T f 


::::::::::::}::::i:::::::::::: 


::::::::::::{::::t:::: ::::::: 




::::::::::::::::"""-:::« 


EEEEEEE=EEEEEEEEEEEE~~EE~£ 



is the slope of the chord PiP« 



approaches P x along the curve. 

This limiting position of the chord 

is the tangent at P x ; this line can be shown in the case of 

the ellipse (or any curve given by an equation of the second 

degree) to cut the curve in two coincident points at (x lt yi) 

and in no other point. 

The method is applied in parallel columns 

to a general problem and to a particular problem. 

Tangent to the ellipse, 

t. + f = l 
a 2 b 2 

at Pi(x h t/jl) on the ellipse. 

Take the second point 

P 2 (xi + h, y t + k) on curve P 2 (3 - 

Substituting, 

6 2 (^ + h) 2 + a 2 (y x + k) 2 - aW = 0. 

4(3 + h) 2 + 25(| + k) 2 - 100 = 0. 

bW + 2 bV^ 4- b°-h 2 4- a 2 y\ + 2 a 2 ky, 4- « 2 ^ 2 - « 2 & 2 = 0. 
But &V 4- « V - a 2 & 2 = 0. 

36 + 24 7t 4- 4 /i 2 4- 04 4- 80 k + 25 fc 2 - 100 = 0. 



25 4 

at A(3, f) 



A, -| 4- k) on curve 



302 



UNIFIED MATHEMATICS 



Subtracting (and canceling) 
2 tfhXi + b 2 h 2 +2 a?ky x + a 2 ft 2 = 0. 24 ft+4 /* 2 +80 ft + 25 ft 2 =0. 
.-. ft(a 2 ft+2 o?y x )=-h{2 Wx^bVi). ft(80+25 ft) = -ft(24+4 h). 
k_ 2b 2 x l + b 2 h ft 24 + 4/i 



2 a % y i + a 2 ft 



A 



80 + 25 ft 



The chord PlP 2 is given by : 

Since P 2 is on the curve, the chord equation may be written : 
2 b 2 x t + b 2 h 



y -2/1 = 



2 a 2 ^! + a 2 ft 



/ x 8 24 + 4 h ( on 

(ic — a?i). w = ! (x — 3). 

^ V ^5 80 +25 ft V ; 



Let P 2 approach P x along the curve ; h and k both approach 
0, ?.e. can be made just as small as you please. Thus if h is 

made .01 in our numerical prob- 
lem, k will be about — .003, which 
can be obtained by solving the 
quadratic 

25 ft 2 + 80 k + 24 h + 4 h 2 = 0, 

for k. It is evident that the con- 
stant here may be regarded as 
24 h + h 2 and that as h = 0, this 
constant approaches zero, and one 
root of k approaches zero. 

Note that the second value of 
k approaches — 2 y lf and corre- 
sponds to the fact that the given value of x, x 1 + h i is the 
abscissa of two points, on the upper and lower parts of the 
curve, respectively. 

Since h and k both approach zero, 

2Wx 1 + b% = 2b*x 1 24 + 4ft =24 

and and 

2flfy!+ a 2 k = 2a?y x - 80 + 25 ft = - 80 



• - 1 




'/- 










i 












ffl* 








t\F«r 


: J_ 


rf~l{~k\-^L .rifct:)" 






vx^\^)y 








: S* ~ t 


f i 




f--i 




t X 




t 












i 








I 








t 












a: 








± 




± 



- is the slope of the chord P1P2 
h 



THE ELLIPSE 303 

and the slope of the chord approaches more and more nearly, 
and as near as you may please to make it, by taking h (and 
thus Jc) small enough, 

-2b 2 x 1 24 



2a 2 y x 80 

Hence the chord approaches a limiting position, given by 
2b 2 x u v 8 24, 

»-*— isil<— *>• */-5=-8o (a; - 3 >' 

which may be written 

&X& + a*y<y = a^ + Matf y - 1.6 = - .3(a - 3) 



a 2 5 2 . 



or 2/ - 1.6 = - .3(x - 3). 



btxjx + a^ - « 2 6 2 = 10 </ + 3 z - 25 = 0. 

or ^4-^=1. 40?/ + 12* -100 = 0. 

a 2 b 2 u 

By precisely this method, step for step, the tangent to 

Ax 2 + By 2 + 2Gx + 2Fy + C = 0, is found to be 
Ax x x + By x y -f £(* + a^) + F(y -f ^)+ 0= 0. 

PROBLEMS 

1. Find the tangents of slope -+- 2 and of slope — 3 to each 
of the ellipses in problem 6 of the preceding set of problems ; 
the five problems, tangents of slope + 2, should take not to 
exceed 30 minutes ; note that after substituting 2 x -f k for y it 
is better procedure, surer and quicker, to combine terms by in- 
spection rather than to expand each binomial before combining. 
This means to pick out the terms containing x 2 , for example, 
and write the sum of these coefficients directly. 

2. Draw at least three of the tangents of slope + 2 in the 
preceding exercise, and three of slope — 3, each to its conic 
as previously drawn. Find the point of tangency algebraically 
and graphically. 



304 UNIFIED MATHEMATICS 

3. Without plotting the ellipse itself plot 12 tangents of 
slope 0, 1, 2, 3, 4, 5, 6, and 10, and of slope — 1 — 1, — 3, and 

- 6 to the ellipse 9 x 2 + 25 y 2 = 900 ; note that these give a 
fair outline of the ellipse. 

4. In each of the ellipses of problem 8, page 297, find the 
tangents at the point whose ordinate is 2, employing the 
results of problem 10 of the same set, and using the formula, 
Axjx + By x y +Q(x + x 1 ) + F(y + y^+G=-0. 

5. Derive by the method outlined in section 11 of this 
chapter the tangent to the ellipse 9 x 2 + 25 y 2 = 900 at the 
point (8, 3.6) which is on the curve ; at (— 6, 4.8) on the curve. 

6. In the ellipse 9 x 2 -f 25 y 2 — 900, verify that the perpen- 
dicular from the focus upon any tangent to the ellipse meets 
it on the major auxiliary circle. Note that the converse is also 
true. This gives a method for drawing the tangent to an 
ellipse from a point outside the ellipse ; explain. 

12. The tangent to an ellipse at a point on the ellipse con- 
structed from the tangent to the auxiliary circle.- — Let (x h y x ) 

be any point on the ellipse ; the tangent is — + ^ = 1 : the 

a 2 b 2 

a 2 
^-intercept, x i} of this* tangent is — , obtained by solving 



.)• 



^L_ _j_ Vm. — i as simultaneous with y = 0. Evidently x i — — 
a 2 b 2 x x 

depends only upon x x and a, not involving b or y x . Hence this 

value would be unchanged if b were taken equal to a. The 

tangent to the major auxiliary circle x 2 + y 2 = a 2 at (x lf y 2 ) on 

the circle is x ± x + y 2 y = a 2 , and the intercept of this tangent 

a 2 

on the a>axis is also — . This gives the following rule for 

drawing a tangent to an ellipse at any point on the ellipse : 

Construct the major auxiliary circle to the given ellipse; find 
the point P 2 on the circle having the abscissa of the given point; 
at the point P 2 construct the tangent to the circle, cutting the 



THE ELLIPSE 



305 



X-axis at T; the line joining T to P ± on the ellipse is the tangent 
to the ellipse. 




Tangent to an ellipse constructed from the tangent to the auxiliary circle a 
the corresponding point 

Xote that even if b is greater than a, this construction gives 
the tangent, but the circle x 2 -\-y 2 =a 2 is then the minoi- 
auxiliary circle of the given ellipse. 

13. The tangent to an ellipse bisects the angle between the 

focal radii to the point of tangency. — Let ^ + ^ = 1 be the 

a 2 b 2 

tangent at (x lf y$ which intersects the cc-axis at T f — , ). 



OT=-; FoT= c ^+ae = -(a + ex l ). 
x x - :,\ x x x 

F l T=--ae = ~(a- ex,). 



Hence, 



F 2 T = a + ex 1 == P l F 2 
F X T a -ex, P^ 



306 



UNIFIED MATHEMATICS 



.-. P X T bisects the exterior angle of the triangle F^P^, 
since it divides the opposite side into segments proportional 
to the adjacent sides. The normal bisects the interior angle 
between the two focal radii ; if the normal be drawn, each focal 
radius makes the same angle with it. 




Tangent to an ellipse constructed from the focal radii to the point of 
tangency 

Any ray of light or sound striking a reflecting surface is 
reflected in the plane of the normal to the surface and the 
original ray in such a way as to make the angle of incidence 
(i.e. between normal and original ray) equal to the angle of 
reflection. Hence rays starting from F x in our figure con- 
verge at F 2 . This is the principle of " whispering galleries," 
in which the rays of sound starting from a point F 1 converge 
at another point F 2 , making audible at F 2 whispers at F x ; at 
intermediate points the conversation may not be audible as 
there is no reenforcement by convergence. 



PROBLEMS 



1. In the ellipse 9 x 2 + 25 y 2 = 900, give the two graphical 
methods for drawing the tangent at the point (6, 4.8) on the 
ellipse. 



THE ELLIPSE 307 

2. In the . ellipse. 25 x 2 + 9 \f — 900, give the graphical 
methods for drawing a tangent at the point (—4.8, 6). on the 
ellipse. 

3. Construct the three ellipses : 

.*- + £=» 1. 

100 36 



i- 



+ -11 = 1. 

100 100 

*_.£_ = ! 

100 144 

Draw the tangent to each of these ellipses at the point 
whose abscissa is -f 6 ; find the equation of each of these 
tangents and prove that they intersect on the x-axis. 

4. Find the equations of the two focal radii to the point 

x 2 v 2 

(6, 4.8) on the ellipse \- *— = 1 : find the bisectors of the 

V ' ; F 100 36 

angles between these focal radii ; find the bisector of the angle 

which does not include the origin, and prove that it coincides 

with the tangent at (6, 4.8) to the ellipse. 

5. If an elliptical arch is to be in the form of the upper 
half of an ellipse, find the equation and plot ten points, given 
that the width of the arch is to be 100 feet and the height is to 
be 40 feet. 

6. If the preceding arch is to have the dimensions as given, 
but is to be constructed as the upper quarter of a vertical 
ellipse, find the equation of the curve. Note that you have a 
point (50, 40) which is to satisfy the equation of the curve 
which can be written with only the denominator of x 2 as un- 
known. Compare this arch with the preceding one as to beauty 
of design. 

7. Find the lengths of the ten vertical chords of the arch 
in problem 5, dropped from the tangent at the top of the arch 
and equally spaced horizontally. 



308 UNIFIED MATHEMATICS 

8. By the method of article 13, find the tangent to the 
curve given by the equation, xy = 15, at the point (3, 5) on the 
curve. 

9. By the method of article 9, find the tangent of slope — 2 
to the curve given by the equation, xy = 15. 

10. Write the equations of the three ellipses of problem 8, 
page 297, in parametric form. 

X iP" 

11. In the ellipse, — — — -f- ^— =1, find where lines from 

(50, 30) inclined to the horizontal axis at angles of 15°, 30°, 
45°, and 60° respectively, cut the ellipse. Eind the lengths of 
these lines from (50, 30). See problem 5. 

12. If in the ellipse of problem 6, supporting chords are 
drawn diagonally between parallel vertical chords, computed in 
problem 7, each from the upper point of the right-hand chord 
to the lower point of the left-hand chord (on the right side of 
the ellipse), compute the lengths of these chords. 




From Tyrrell's Artistic Bridge Design 
Alexander III Bridge in Paris 
A parabolic arch ; span, 107.6 m.; rise, 6.75 m.; width, 40 m. 

CHAPTER XIX 



THE PARABOLA 

1. Definition. — The ellipse has been 
defined (page 289) as the locus of a 
point which moves so that its distance 
from a fixed point, the focus, is in a 
constant ratio less than one to its dis- 
tance from a fixed line, the directrix. 
If this constant ratio is taken equal to 
one, the curve generated by the mov- 
ing point is called a parabola. 

PF = e • PZ, e < 1, defines an ellipse. 

PF = PZ defines a parabola. 

PF = e • PZ, e > 1, defines a hyperbola. 

2. Equation of the parabola. — 

Through the focus draw the perpendic- 
ular FQ to the directrix ; take this line 
as 3>axis. Take FQ, which is constant, 
as 2 a. On the axis chosen only one 

309 



IP 




y 






Z r -+- 




1 




J 




Jt 






.^~ 


_x 




J 




+ 




r 


V 4). TP-j/ 




Hk.% Al 




■^ * 




V 
















~AL 


Ap' 























































F, the focus; ZZ", the di- 
rectrix 

A point equidistant 
from .Fand ZZ' moves on 
a parabola. 



310 UNIFIED MATHEMATICS 

point is found which is on the given curve ; the mid-point 
0, dividing the segment QF in the ratio 1 to 1, is such that its 
distance from F, the focus, equals its distance from the 
directrix. Through take a perpendicular to OX as the 
2/-axis. Evidently F is the point (a, 0), and the directrix is 
the line x -f- a = 0. 

Take P(x, y) any point which is on the given curve, i.e. any 
point equally distant from F and from the line, giving 

PF= PZ. 



PF = V(sc — a) 2 -+- y 2 , distance between two points. 
PZ = x + a, distance from a point to a line. 

Note — ^— gives the distance as negative ; but it is not 

necessary to take account of the sign as in the simplification 
this expression is squared. 
Equating, PF = PZ, gives 



V (# — a) 2 + y 2 = x -\- a. 

x 2 — 2 ax + a 2 + y 2 = x 2 + 2 ax +^a 2 . 

2/ 2 = 4 a#. 

3. Right focal chord. — When x = a, y —±2 a, giving the 
total length of the right focal chord as 4a; the coefficient of 
x in y 2 = 4 ax represents the length of the right focal chord. 

4. Geometrical properties. — The curve is symmetrical with 
respect to the a?-axis, for, assigning any value to x, you find for 
y two values ± V4 ax ; numerically equal but opposite in sign, 
or lying symmetrically placed with respect to the OX-line, 
which consequently is here the axis of the curve. 

The ?/-axis, x = 0, is tangent to this curve since, solving, 
?/2 — 4 aX) 
x = 0, as simultaneous, 

gives 2/ 2 = 0;?/ equals zero twice, or the two points of inter- 
section of x = with y 2 = 4 ax are coincident. The point 



THE PARABOLA 311 

(0, 0) is the point on the axis of symmetry, y = 0, which 
corresponds to itself ; this point is called the vertex. The 
line y = cuts the curve in only one finite point, given by 
?/ = 0, and x = ; the other point of intersection of y = 
with the curve is at an infinite distance. 

G-iven a as positive, negative values of x lead to imaginary 
values of y. Hence all points on the curve lie to the right of 
the tangent at the vertex. As x increases without limit, y 
increases also without limit. This curve extends, we may 
sa} T , to infinity. In plotting a parabola, the vertex, the ex- 
tremities of the right focal chord, and at least two other 
points, should be plotted. 

The quantity y 2 — 4 ax is evidently negative for points in- 
side the curve, zero for points on the curve, and positive for 
points outside the curve. 

5. Finite points and the infinitely distant point on the parabola. 
— To plot carefully y 2 — 8 x, note that 4 a = 8, whence a = 2 -, 
indicate the vertex and, 2 units to the right, the focus ; 4 
units (2 a) above and below the focus locate the extremities 
of the right focal chord which has the length 4 a ; take values 
of x at appropriate distances from the vertex and focus to 
give the portion of the parabola desired ; in y 2 = 8 x, x = 0, 1, 
2, 3, 4, 6, and 8 give sufficient points to plot the curve for our 
purposes. 

The parabola y 2 = 8 x is intersected by the line y = x at 
(0, 0) and at (8, 8) ; y = -J x cuts this parabola at x = and at 
x = 32 ; y = .1 x cuts at x = 800, y = 80. These values are 
obtained by solving the equation y 2 = 8 x as simultaneous with 
each of the linear equations. 

_, , . ?/=-01 x 

feolvin s> f=%* 

gives .0001 x 2 = 8 x, x = or x — 80,000. If one centimeter is 
taken as 1 unit, y = .01 x cuts the parabola y 2 = 8 x at the 
vertex and at a distance of 80,000 cm., nearly \ mile, from the 
vertex. As the line y = mx moves nearer and nearer to the 



312 



UNIFIED MATHEMATICS 




A line parallel to the axis of a parabola cuts it in a point infinitely distant 

axis y = 0, the second point of intersection moves off farther 
and farther, without limit. This is the meaning of the ex- 
pression that the axis of the parabola, and by similar reason- 
ing any line parallel to the axis, " cuts the curve at an infinite 
distance." 

The methods given above for plotting y 2 = 8 x apply to any 
parabola y 2 = 4 ax. 



PROBLEMS 

Plot the following parabolas carefully : 

1. y 2 = 4 x, from x = to x = 8. 

2. y 2 = x, from x = to x = 10. 

3. y 2 = 12 x, from x = to x = 12. 

4. y 2 = T !g- x, from x = to x = 100. 



THE PARABOLA 



313 



5. s = 16 t 2 , or t 2 = Jg s, taking OS as the horizontal axis 
and taking J inch to represent 10 units of s (distance in feet) 
and \ inch to represent 1 unit of t (time in seconds). This 
gives the distance fallen from rest in time t by a freely fall- 
ing body. 

6. Find the intersections of y = x, y == ^ x, and y = ^x 
with the curves of problems 1, 2, and 3. 

7. Solve s = 800 t with s = 16 t 2 ; s = 800 t is the space cov- 
ered by a body moving with uniform velocity 800 units per 
second. What is the physical meaning of the values obtained 
for the point of intersection ? 

8. A mass rotated on a cord exerts a force of tension on 
m • v 2 



the cord, F 



Given m = 1 pound, and r = 10 feet ; 



draw the graph for velocities of 1 to 100 feet per second, taking 
F on the horizontal axis. Compute the corresponding number 
of revolutions per minute for v = 10, 20, 50, and 100 feet per 
second. "What is the relation between 
v and n, where n is the number of 
revolutions per minute ? Indicate on 
the vertical scale a second scale giv- 
ing n. 



6. Geometrical interpretation of 

y 2 = 4 ax. 

The equation y 2 = 4t ax may be inter- 
preted geometrically as follows : 

MP 2 =±a-VM, 



S 



M. 



m 



the square of the perpendicular from 
any point on a parabola to the axis is 
equal to the rectangle formed by the 
length cut off on the axis from the ver- 
tex by the perpendicular, with a constant line of length 4 a, 
the length of the right focal chord. 



The square on PM equals 

the rectangle with VM 

and RFC as sides 



314 



UNIFIED MATHEMATICS 



7. Standard and limiting forms. — Given that the axis of a 
parabola is parallel to one of the coordinate axes, the relation, 

MP 2 = 4 a ■ VM 

leads to 2 (or 4) stand- 
ard forms of the 
parabola. > 

Thus in the upper 
figure a parabola is 
drawn, having V(— 2, 
5) as vertex, 4 a = 4, 
axis parallel to the. 
x-axis and opening 
to the right ; the re- 
lation 

MP 2 = 4 a • VM 

leads to 
(y-5y = ±(x + 2). 

Were V the point 
(h, k), VM would be 
x — h since it is the 
distance from a point 
whose abscissa is h to 
a point whose ab- 
scissa is x ; similarly MP is y — k ; now when the curve 
opens to the right VM is positive ; hence the equation is 
(y — k) 2 = 4 a(x — h), with a positive. Were the axis parallel 
to the x-axis, but the curve opening to the left, the equation 
would be : 

(y — k) 2 = 4 a(x — h), with a negative. 

Similarly the curve on our figure which opens down is given 
by the equation (x — 3) 2 = — S(y — 2). 

In the general case, (x — h) 2 = 4 a (?/ — k) has V(h, k) as 
vertex ; the axis is x — h = and is parallel to the y-axis ; 




A horizontal and a vertical parabola 



THE PARABOLA 



315 



the curve opens up when a is positive and down when a is 
negative. 



gm- 



(y - k) 2 = 4 a(x - h) 




mm 



GEE 



: =J3w= 



ffiSSK 



(x-hy = 4a(y-k) 



Standard forms of the parabola equation 

As a approaches in y 2 = 4 ax, the parabola approaches 
more and more nearly to coincidence with the x-axis ; two 
coincident straight lines constitute a limiting form of the 
parabola. As a becomes larger, the parabola approaches the 
?/-axis. 

8. Tangent of slope m. 

y- — 4 ax. 

y = mx -f- Je. 

m 2 x 2 + 2 Jcmx -f- Jc 2 — 4 ax = 0. 



_ (2 a - km) ± V(2 a - km) 2 - m 2 k 2 
x — 1 



1)V 



whence, since A = 0, 



m 



y = mx -\ — -is the tangent of slope m to y 2 = 4 ax ; 



is the point of tangency. 



9. Tangent from an external point. — For any given point 
O^'d 2/i) outside the curve two values of m will be found for 

vhich y = mx + -• will pass through (a^, y x ) ; hence there 

7/1 



316 



UNIFIED MATHEMATICS 



are, in general, two tangents which pass through a given point 
outside the curve. XlM 2 _ y{)n 4. a = o, 



m = 



2/i±V?/i 2 -4a3V 
2x x 



For points inside the curve, y x 2 — 4 ax 1 is negative, and there 
are no tangents. 

10. Tangent at a point (x 1} y x ) on the parabola. — By the 

method of article 9 of the preceding chapter the tangent to 

the parabola, if = ± ax, at (x 1} y t ) 

on the curve is found to be 

y 1 y = 2a(x + x 1 ). 

Similarly the tangent to 

By* + 2 Gx + 2Fy + C = 
is found to be 

By,y + G(x + x x ) + F(y + yi ) + C=0; 
and with the x 2 term present, Ax x x replaces in the above ex- 
pression the term By x y. 

11. Illustrative example. — Put in standard form and plot 
carefully 4 y 2 _ 12 y 4 6 x - 11 = 0, 

4 z (jj2_s y )=-6a + ll. 

% 2 -32/ + |) = 

-6x + ll + 9 = -6a; + 20, 

completing the square inside the 

parenthesis. 

%-f) 2 = -^-- 2 /) 
= -6(x-i£-). 

(2/-|) 2 = -f(^-- 1 s -). 
F(JjM); type a-; axis y-# =0; 
4 a = — f ; a = — -|. 
Plot F, F, RFC, and the further point where y = \ ; 




«-■ ¥ = 



— _ 



35 



=*■ 



# = 



THE PARABOLA 317 

Draw a smooth curve tangent to x = -L -, at V(~^- } f ) through 
the points which are plotted ; here it would be well to find 
from the original equation the intercepts on the axes. 

PROBLEMS 

Put in standard form and plot : 

1. 4*/ 2 -12?/ + 6a;-ll=0. 5. (y - 3) 2 = $x + 11. 

2. 4,x 2 -12x-6y-ll = 0. 6. y 2 = 6x + ll. 

3. ±x 2 -12x + 6y-ll = 0. 7. y 2 = ^x. 

4. if-6y-Sx-5 = 0. 

8. Solve graphically, to 1 decimal place, 

X 2 + y 2 = 25 

2/ 2 = 8 x', by drawing both graphs to the same axes. 

Put in standard form and plot the equations obtained in the 
two following problems : 

9. The formula for the height of a bullet shot vertically 
u] 'ward with a velocity of 800 feet per second, s = 800 1 — 16 t 2 . 

10. The formula for the time of beat, in seconds, of a pendu- 

lum is t 2 = — • I, taking g = 980 and I measured in centimeters ; 

taking g = 32, I must be measured in feet. 

Compute corresponding values of t and I by logarithms, cor- 
rect to 3 significant figures. Would the diagram be changed 
if g is taken as 982 instead of 980 ? At sea level on the equator 
g = 978.1 cm./sec. 2 ; at Washington, 980.0 ; at New York, 980.2 ; 
at London, 981.2 ; or in feet/sec. 2 32.09, 32.15, 32.16, and 32.19, 
respectively. 

11. The Hell-Gate steel arch bridge in New York is one of 
the largest arch bridges in the world. See the illustration, 
p. 353. The lower arc of the arch is a parabola, 977.5 feet as 
span and 220 feet as height of the arch. Write the equation 
of the arc, taking as :oaxis the tangent at the vertex of the 
parabola and as ?/-axis the axis of the parabola. Compute 
4 a to 1 decimal place. The roadway is 130 feet above the 



318 UNIFIED MATHEMATICS 

base of the arch ; compute the length of the roadway between 
the parabolic arcs. There are 23 panels or openings, spaced 
42.5 feet apart at the centers ; compute the vertical lengths to 
the roadway from the arc of the parabola, also to one decimal 
place. Compute the approximate length of the parabolic arch it- 
self by computing the lengths of the twenty-three chords on the 
parabola ; note that only 12 computations are necessary ; do not 
carry beyond tenths of a foot, as hundredths would have little 
significance. Locate the focus and the directrix of this parabola. 

12. Engineers use the following method for constructing a 
parabolic arch ; show that it is correct. Suppose that it is 
desired to construct a parabolic arch of width 100 feet and 
height 30 feet ; a rectangle 50 by 30 is drawn and the right- 
hand side is divided into 10 (or n) equal parts which are joined 
to the upper left-hand vertex of the rectangle (and parabola) 
by radiating lines ; the upper horizontal side is also divided 
into 10 (or n) equal parts and ordinates are drawn at these 
points ; corresponding lines intersect at points on the parabolic 
arch desired. Of the lines drawn the second (op rth) ordinate 
to the right of the vertex corresponds to the second (or rth) 
radiating line drawn from the vertex to the second (or rth) 
point of division from the top, on the right-hand side. 

13. If an ordinary automobile headlight reflector is cut by 
a plane through its axis the section is a parabola having the 
light center as focus. If the dimensions of the headlight are 
10 inches in diameter by 8 inches deep, locate the focus. 

14. Locate the focus of a parabolic reflector, 6 inches in 
diameter and 4 inches deep ; 5 inches deep ; 6 inches deep. 

15. The cable of a suspension bridge whose total weight is 
uniformly distributed over the length of the bridge takes the 
form of a parabola. Assuming that the cable of the Brooklyn 
bridge is a parabola, which it is approximately, find the equa- 
tion in simplest form ; the width between cable suspension 
points is about 1500 feet and the vertex of the curve is 140 feet, 
approximately, below the suspension points. 



THE PARABOLA 319 

16. The Kornhaus Bridge over the Aar at Berne, Switzer- 
land, has for central arch a parabola ; the span is 384 feet and 
the height of the arch is 104 feet. If there are vertical 
columns spaced 24 feet apart, determine the length of these 
columns, assuming that the roadbed is 30 feet above the vertex 
of the parabola. If the floor of the roadbed is on a 2.7 per 
cent grade, determine the difference in elevation between the 
center of the bridge and the ends. 

17. The parabolic reflector at the Detroit Observatory, 
University of Michigan, has a diameter, which corresponds to 
arch span, of 37.5 inches ; the focal length of the mirror is 
19.1 feet, from vertex to focus. Determine the height of the 
arch (or the depth of the reflector) ; determine the equation 
of the parabolic curve. The rays from a sun or star which 
strike this surface parallel to the axis of the parabola converge 
at the focus. What is the slope of this mirror at the upper 
point of the mirror ? At the point whose abscissa is i inch ? 

18. To draw a tangent to a parabola from an external point 
you can proceed as follows : take the external point as center, 
the focal distance as radius, and describe an arc cutting the 
directrix ; from the point of intersection draw a line parallel 
to the axis ; the intersection point with the parabola is the 
point of tangency. Prove this method. 

19. Find the tangents of slope + \ and — .3 to each of the 
parabolas in exercises 1 to 7 ; time yourself. 

20. Find the tangent to each of the parabolas in exercises 
1 to 7 at the point on each parabola whose abscissa is -f 2 ; 
the exercise should be completed within thirty minutes. Show 
the geometrical method of working one of these problems. 

21. Find the tangents to the parabola in problem 1 from 
the point (2, 10) outside the parabola. Describe a geometrical 
method of working this problem after the graph of the parab- 
ola is drawn. 

22. Plot to scale with the dimensions given the fundamental 
parabola of the Alexander III Bridge. 



CHAPTER XX 



THE HYPERBOLA 

1. Definition and derivation of the equation. — (See ellipse, 
page 289, and parabola, page 309.) 

PF=e-PZ, e>l. 

Take FX perpendicular to D'D as the aj-axis, intersecting 
the given directrix at Q. 




Let A and A' divide the segment QF internally and ex- 
ternally in the ratio e (f in the figure). The mid-point of 

320 



THE HYPERBOLA 321 

AA l , is taken as the origin and the perpendicular through 
this point to X'X as the t y-axis, OA = OA' = a. 
Precisely as in the ellipse, 

AF=e AQ, 
A'F=e-A'Q', 
whence AF+ A'F = e • (AA 1 ). 

AF+OA'+OF=2ae, 
2 0F=2ae, 
OF=ae, 

and, by subtraction, OQ = -• 

e 

F is (ae, 0) ; D'D is a? - - = 0. 
The relation PF= e ■ PZ gives the equation, 
V(z - ae) 2 + J/ 2 = ? (» - - 

a;2 _|_ a 2 g 2 _|_ y 2 = e 2 x 2 + a 2 # 
aj2(l _ e 2) + y 2 = a 2(l _ g2). 

Up to this point the work is practically identical with the 
work in the case of the ellipse ; here, however, 1 — e 2 is nega- 
tive, since e > 1. Hence we write this equation, 

0;2( e 2 _ 1) _ y 2 _ a 2( e 2 _ 1). 

* 2 y 2 =1 

a 2 a 2( e 2 _ 1) 

Let 5 2 = a 2 (e 2 - 1). 

■*? _ yl — 1 

a 2 6 2 ~ 

v-2 t/2 

2. Geometrical properties of the hyperbola, — = 1. — 

a 2 ft 2 

Since y and — y lead to the same values of x, the curve is 

symmetrical with respect to the #-axis ; since x and — x lead 

to the same values of y, the curve is symmetrical with respect 



322 



UNIFIED MATHEMATICS 



to the y-Sixis. Hence the intersection of the two axes is the 
center of the curve. 

b 



Solving for y, 



y 



± - Vx 2 - a 2 
a 



this expression shows that the . curve is symmetrical with 
respect to the o?-axis, for any value of x gives two values of y 
equal in value but opposite in sign. Since any value of x 




Symmetry of the hyperbola 



numerically less than a gives imaginary values of y, the curve 
lies wholly outside the region bounded by x -+- a. = and 
x — a = 0. When x=±a, y = 0; these are self-corresponding 
points on the horizontal axis of symmetry ; these points are 
called the vertices. When y = 0, x is imaginary ; the vertical 
axis of symmetry does not intersect the curve. 



Solving for x, x = ±- V?/- 
b 



b 2 ; the curve is symmetrical 

with respect to the y-axis ; every real value of y gives two 
corresponding real values of x, symmetrically placed with 



THE HYPERBOLA 323 

respect to the ?/-axis ; the vertical axis of symmetry does not 
cut the curve in real points. As y increases in value, without 
limit, so do the two corresponding values of x increase in 
value, numerically without limit. 

Since there is this vertical axis of symmetry it is evident, 
precisely as in the ellipse, that there is a second focus, 

F 2 (— ae, 0), and a corresponding directrix, x + — = 0. 

e 

The axis which cuts the curve is called the principal axis ; 

the other axis is called the conjugate axis. The lines x= ± a 

are tangent to the curve at the vertices. See page 310. 

3. Right focal chords. — The foci are the points (ae, 0) and 



(— ae, 0) ; when x = ± ae, y == ± - VaV — a 2 = ±b Ve 2 — 1 ; 

a 

b 2 
but since b 2 = a 2 (e 2 — 1), these values of y equal ± — ; each 

2 b 2 

right focal chord is of length , and is constructed by 

a 

erecting at the focus lines perpendicular to the principal, • or 

b 2 
transverse, axis of length — on each side of the axis. 

a 
The foci are at a distance ± ae from the center ; now 
b 2 = a 2 (e 2 — 1) gives ae = Va 2 + b 2 , which is the length of the 
diagonal of a rectangle of sides a and b. 

4. Finite and infinitely distant points on the hyperbola. — 

Given to plot the hyperbola 2— = 1, .note that a 2 = 16, 

1 J1 16 49 

6 2 = 49; 49 = 16(e 2 -l), whence e 2 - 1 = ff , e 2 = ff, and 

ae = V49 + 16 = V65; 5? = ^?. 
a 4 

It will be found convenient to draw the rectangle having 
O as center and extending 4 units to the right and left of O 
and 7 units above and below. The half-diagonal of this 



324 



UNIFIED MATHEMATICS 



rectangle has the length V65, and may be used to locate on 
the principal axis the two foci. 



m$> 




Asymptotes and one branch of the hyperbola — — — = 1 

16 49 



The line y = x cuts the curve at — _~ = 1, 33 ^ =1; 

16 49 ' 16 x 49 

, = ±-^ = ±^^=±488. 
V33 . 33 

The line y = mx cuts the curve in two points, whose ab- 
scissas are given by 



x L m l x l 
16~~^9~ 



= 1, or x 2 = 



49 x 16 
49 - 16 m 2 



; x = ± 



28 



V49 - 16 m 2 



THE HYPERBOLA 325 

As 16 m- approaches nearer and nearer to 49 these two 
points of intersection move farther and farther off; when 
49 — 16 m 2 = 0, m = ± f , the two points of intersection of each 
of these lines with the hyperbola move off to an infinite dis- 
tance : the lines y = \x and y = — \x are called asymptotes of 
this hyperbola, intersecting the curve in two coincident points 
both at an infinite distance. 

In the hyperbola - — — = 1, the two lines y — ^x and 
a- b 2 a 

y = Xj or : — " — and — -f- i - = are asvmptotes ; note 

a a b a b 

that i *] [- + -] gives the left-hand member of the equa- 

\a b) \a bj 

tion of the hyperbola, when the right hand is unity. 
5. Illustrative problems. — Plot the hyperbola 

16 49 

The rectangle of sides 8 and 14, parallel to x- and ?/-axes respectively, 
is plotted with its center at the origin. As noted above, the diagonals 
of this rectangle give the distance from the center on the transverse axis, 
here horizontal, of the foci ; the diagonals extended are the asymptotes, 
and to these lines the curve approaches more and more nearly as the 
curve recedes towards infinity. The right focal chords have the total 

lensrth - — ; plot — vertically above and below the foci. Take x = 6, 

a 4 

this gives another point between vertex and right focal chord ; 

£ = 1" 1 ' *=±7V06 = ±7(l.ia)==±7.84j 

49 lo 

take x = 10, y - ± 7 y/hM = ± 7(2.29) = ± 16.03. 

Plot also the symmetrical points. 

PROBLEMS 

1. Plot the hyperbola — - ^- = 1. 

J1 49 16 

2. Plot the hyperbola — ^- = 1. 

Jk 100 100 

This type of hyperbola, a = b, is called an equilateral or rec- 
tangular hyperbola. Why ? 



326 



UNIFIED MATHEMATICS 



3. Plot the hyperbola ^-— ^- = 1 ? computing values re- 

• Ok Ou 

quired to one decimal place. 

4. The equation of the hyperbola - — ^- = 1 may be put in 



parametric form, 



x = a sec 
y = b tan 



Noting that sec is . find by using logarithms five 

cos 6 

points on each of the above hyperbolas. Check on the graphs 

drawn. What is the geometrical significance of ? 

5. Find the equations of the asymptotes of each of the pre- 
ceding hyperbolas and find to V the angle of inclination to 
the horizontal axis. 

6. Compare the right-hand branch of the hyperbola 

— — -- = 1, with the parabola y 2 — — (x — 4) ; this parabola 
-Lo y -Lo 

has the same vertex and passes through the extremities of the 

right focal chord. Prove this. Do these curves coincide in 

other points ? 




Focal distances to Pi (xi, yi) on the hyperbola 



THE HYPERBOLA 



327 



6. Difference of the focal distances constant. — Designate the 
right focus by F x and the corresponding directrix by DD', 
and the left focus by F 2 , having D 2 D' 2 as the corresponding 
directrix. Then the focal distances PF 1 and PF 2 are ex l — a 
and ex x + a, respectively ; the difference, PF 2 — PF lF = 2 a, 
is constant. 

The hyperbola may be defined as a curve generated by a 
point which moves so that the difference of its distances from 
two fixed points is constant. 

7. Standard forms of the equation of the hyperbola. — 

(x-ny (y-ky _ ± 

a 2 6 2 

(y-ky {x-hy _ 1 

a 2 6 2 

Precisely as in the ellipse, article 6, Chapter 18, the equation 

of the hyperbola ^ - 1^ = 1, may be interpreted 



b 2 
OM 2 



MP' 



= 1. 



OA 2 OB 2 

Eor any hyperbola whose axes of symmetry are parallel to the 
coordinate axes we obtain, from this relation, the equations 

(x-hy (y-W = 1 (y-fc) 2 (x-hy = 1 

a 2 V- and a 2 b 2 

for a horizontal hyperbola, for a vertical hyperbola. 





Horizontal hyperbola 



Vertical hyperbola 



328 UNIFIED MATHEMATICS 

8. Conjugate hyperbolas and limiting forms of the hyperbola 
equation. — Given any hyperbola, 







X 2 


b 2 ' 


the lines 




X 2 

a 2 ' 


-t =0 
b 2 


represent 


the 


asymptotes 5 the equation 






X 2 


-£ — 1 



a 2 b 2 

represents a vertical hyperbola about the same rectangle and 
having the same asymptotes. Any two hyperbolas so related 
are called conjugate hyperbolas. 
Illustration. 

(J^I_(JL±2Y= ! and (*L^i 2 _ fe±2)_ 2 = _ i are con . 

16 49 16 49 

jugate hyperbolas ; the second is written in standard form 

(y _j_ 2V (x — 3) 2 

v// — L ^ '— — 1, wherein the focal distances ± ae from 

49 16 

the center, and the distances ± - of the directrices from the 

e 

center (3, — 2) are obtained, regarding a 2 as 49 and b 2 as 16. 
The asymptotes of these conjugate hyperbolas are given by 

( x Q\2 (y I O) 2 

the equation ^ — ^—^ — '- = 0, or by the equivalent in 

separate factors, 

^z3_^ = 0and En3 + &±2_0. 

4 7 4^7 

(x — 7i) 2 (v — 7c) 2 
The equation ^ '- — ^ '- = 0, representing two real 

straight lines, is a limiting form of the hyperbola equation, 

(x - h) 2 (y + k) 2 A , b 

■^ '- — ^_j — /_ = m. As m approaches zero, - remains con- 

a 2 b 2 ^ F 'a 

stant and the hyperbola approaches more and more nearly the 

two straight lines ^^ ^ ~~ = and x ~ + y ~ = 0. 

a b a b 



THE HYPERBOLA 



329 



9. The equilateral or rectangu- 
lar hyperbola. — The hyperbola 

r' 1 u 2 

- — — = 1 is called an equilat- 

a 2 a 2 

eral hyperbola since b = a ; it is 
also called a rectangular hyper- 
bola as the asymptotes are at 
right angles to each other. 

Since b 2 = a 2 (e 2 — 1), the value 
of e in an equilateral hyperbola 
is V2 or 1.414; for e > V2, 
b 2 >a 2 ; for e<V2, b 2 < a 2 . 





5 a£ 




- \j\ ~^ lkM v 


xs v ,v/fy 


^SV it „"ZZa 


SSIX ^ZZT^ 


- >SZS r „<Z/SX. 


\\ N. -^ Z^Lrli 


\ > *> ^ ^/r> 


^5" :?:o 




^ : s zn f 








Fty 7 Z 0- S c k^1 






_ 1 Z__-,..___5 5I£ _ 




,/z *'' "i^s^r 


*^-<' "-AX 


^^a- \ss 


/'s \S^ 


//■ >^S 


JcP ^v 


Z* '*; 


/ V 


2 S 



The equilateral or rectangular 
hyperbola 



10. Illustrative problem. — Put the equation 
4 x 2 +-16x -9y 2 -lSy-75 = 
in standard form and plot the curve. 

4(x 2 +jta; )-9(y2 + 2y )=75 
4(x 2 + 4x + 4)- 9(y 2 + 2y + 1)= 75 + 16 - 
4( X + 2)2- 9(y + l)2 = 82 
(x + 2) 2 Q/ + 1) 2 



20.5 

The center is at (—2, 
a 2 = 20.5 and a 



V20.5 + 9.11 = V29T6i = 5.44 ; 




9.11 

1) ; the hyperbola is of horizontal type ; 
4.53; 6 2 = 9.11 and 6 = 3.02 ; 
5 2 _ 9.11 . 
a IM' 

further convenient points 
are given by x = 5 ; substi- 
tuting in the original is 
easiest, giving 

9 y 2 + 18 y — 105 = ; 
y 2 + 2 y -ajp = ; 



Upper portion of right-hand branch of the 
given hyperbola 



v=-i ±Vi+i^ 
= -l±K10.68) 
= -l±3.56. 



330 UNIFIED MATHEMATICS 

PROBLEMS 

1. Put the equation 

4 x 2 + 16 x - 9 y 2 - 18 y + 107 = 0, 
in standard form and plot. 

2. Plot one quarter of the hyperbola 

x 2 y 2 =1 
147 2 59 2 

3. Plot a hyperbolic arch, width 200 feet, height 60 feet, 
as part of a rectangular hyperbola. Assume the equation 
y2 — x 2 = a 2 , and note that (100, a + 60) is on the curve. 

4. What limitation is there upon the values of A and B, 
if the equation Ax 2 + By 2 +2Gx + 2Fy + C=0 represents a 
hyperbola ? 

5. Any equation of the form xy — ~k 
or (a x x + b x y + c x ) (a 2 x + b 2 y + c 2 ) = k, 

has for its locus a hyperbola ; the lines obtained by equating 
the left-hand member to zero are the asymptotes. Plot the 
hyperbolas xy = 10 and (x — 3 y)(x — 4 y) = 50. 

6. Put the following equations in standard form, completing 
the square first and reducing to standard form by division. 

a. 4a 2 -92/ 2 -8a + 36?/ = 0. 

b. 3x 2 + 24a-2/ 2 +6?/-43 = 0. 

c. 5cc 2 -17z-102/ 2 + 100 = 0. 

d. 5 a 2 - 12 y 2 - 117 = 0. 

e. 3a 2 -24z-42/ 2 -162/-52 = 0. 

7. Plot the preceding five hyperbolas, choosing an appro- 
priate scale. Plot the extremities of the conjugate axes ; 
plot the rectangle and its diagonals ; plot the extremities of 
the right focal chords ; plot at least one further point, 
properly chosen to give the form of the curve, and its 
symmetrical points with respect to the axes. It is desirable 
to plot at least two of these curves completely ; the remaining 
curves need be sketched only in the first quadrant. 



THE HYPERBOLA 331 

8. Determine a 2 and b 2 to one decimal place in the follow- 
ing three hyperbolas : 

a. 17 x 2 - 43 y 2 = 397. 

b. 5 x 2 - 17 x - 10 y 2 - 35 y = 0. 

c. 70 - 2) 2 - 3(y - 3) 2 = 39. 

9. In the three hyperbolas immediately preceding deter- 
mine ae, — , and - to one decimal place. 

a e 

10. In each hyperbola of problem 8 determine x when y — 2. 

11. Using- the data of the three preceding problems, plot 
the three hyperbolas of problem 8. 

x 2 v 2 

12. In the hyperbola — > — M— = 1. find the coordinates of the 

JF 64 36 ' 

foci. What is the distance of the point whose abscissa is 12 
from each of the foci ? of the points whose abscissas are 10, 11, 
15 ? State the general form for this distance. 

13. Put the following equations in standard form and discuss, 
the curves represented by these equations : 

a. x 2 — 6 x — y 2 — 6 y = 0. 

b. z 2 -6a-4?/ 2 -82/ + 7 = 0. 

c (*-3) 2 (y + 2)2 
25 36 



CHAPTER XXI 

TANGENTS AND NORMALS TO SECOND DEGREE 
CURVES 

1 . The general quadratic in x and y. — The general equation 
of the second degree in x and y is written, 

Ax* + 2 Hxy + By* + 2 Gx + 2 Fy + C~ 0. 

The equations of the circle, parabola, ellipse, and hyperbola 
are special types of this general equation. Since none of 
these standard forms, representing curves of the second degree 
with axes of symmetry parallel to the coordinate axes, have 
an xy term, we find it convenient to discuss the general 
equation, with H=0, or 

Ax* + Bf +2Gx + 2Fy + C=0. 

This represents one of the curves — circle, ellipse, parabola, 
or hyperbola — mentioned above, or some limiting form of the 
same, including pairs of lines and imaginary types. It can be 
shown that 

Ax* + 2 Hxy + By* + 2Gx + 2Fy+C=0 

represents no new curve ; simply one of the above-mentioned 
types turned at an angle to the coordinate axes. 

2. General equation of the second degree represents a conic 
section. — Given a right circular cone, it can be shown by the 
geometrical methods of Euclidean geometry, that the section 
which is made with the surface of the cone by any plane is 
one of the curves above mentioned ; thus a plane parallel to 
the base cuts the cone in a circle, or in a point circle if through 
the vertex. 

332 



TANGENTS AND NORMALS 333 

The cone is conceived as the whole surface determined by 
the straight line elements of the cone produced to infinity. 

A plane which runs parallel to only one of the elements cuts 
the cone in a parabola, or in two coincident lines if the plane 
passes through an element and a tangent to the circular base 
of the cone. 

A plane which cuts all the elements in finite points cuts 
the cone in an ellipse ; this is a point ellipse when the plane 
passes through the vertex of the cone. 

A plane which cuts the cone parallel to the plane of two 
elements cuts it in a hyperbola ; if the plane passes through 
the vertex the hyperbola reduces to two straight lines. 

3. Historical note on conic sections. — The fundamental prop- 
erties of conic sections were discovered by Greek mathema- 
ticians nearly two thousand years before the invention of 
analytical geometry which was perfected by Descartes and 
Eerinat, Erench mathematicians of the seventeenth century. 
A treatise on conies was written by Euclid (c. 320 B.C.), but it 
was entirely superseded a century later by a treatise by Apol- 
lonius (c. 250 b.c) of Perga, whose treatise included most of 
the fundamental properties which we discuss. The proper- 
ties of the parabola connected directly with focus and directrix 
are not included in the eight books (chapters) on conic sec- 
tions by Apollonius, nor was the directrix of the central conies 
employed by him. Pappus of Alexandria (c. 300 a.d.), almost 
the last of the Greek mathematicians of any note, included 
these in his Mathematical Collections. 

The Greek mathematicians were interested in these curves 
for the pure geometrical reasoning involved. That the paths 
of the planets were conies they did not know ; nor did they 
know any practical applications of these conies. However, 
the fact that Greek mathematicians had studied these prop- 
erties made it possible for John Kepler and Isaac Newton 
to establish the laws of movement of the planets in the uni- 
verse in which we live. The men mentioned and Nicolas 



334 UNIFIED MATHEMATICS 

Copernicus, who reasserted the heliocentric theory of the uni- 
verse, were all thoroughly versed in the pure geometry of the 
Greeks ; their new theories were built directly up©n this 
foundation of pure geometry. 

4. Tangent of slope m to a second degree curve. — Any line 
y — mx + k cuts a curve given by an equation of the second 
degree in two real points, or in two imaginary points, or in 
two coincident points. The abscissas of the points of inter- 
section are given by the quadratic in x obtained by substitut- 
ing y = mix -\- k in the equation of the curve ; the two equations 
are solved as simultaneous equations. The condition for tan- 
gency is that the two points of intersection of the line with 
the curve shall be coincident ; this will be the case when the 
two roots of the quadratic in x, i.e. the two values of the 
abscissas of the points of intersection, are equal. 

When the intersections of a line with a quadratic curve are 
given by a linear, instead of a quadratic, equation, the mean- 
ing is that one point of intersection has moved off to an in- 
finite distance. As the coefficient of the square term of a 
quadratic approaches zero one root becomes larger and larger 
without limit ; see page 98. 

Parabola, y 2 = 4 ax, 

y = mx + k. 



Solving, a , = -(fcm-2a)±V4a(q-m^ 

m 2 



For equal roots, or coincident points, k = 



m 



A. y = mx -f — is tangent to y 2 = 4 ax at [ — , — V 
m \m 2 m J 



A 



Ellipse, — -|- — = 1, and the line y = mx + k. 



Solving, x = -^W^aW + y-g) , 
5 6 2 + a 2 m 2 



TANGENTS AND NORMALS 335 



B. y = mx ± ^/a 2 m 2 + b 2 is tangent to the ellipse 



x 2 { V 2 = 1:itXz= j «WqW + b 2 ^ y = 



a 2 b 2 a 2 m 2 + b 2 ^a 2 m 2 +b 2 

For every value of m there are two real tangents to an 
ellipse. Similarly 



C. y = rtix ± Va 2 m 2 — b 2 

is tangent to the hyperbola, 



^ = lat*=± 



a 2 b 2 ^/a 2 m 2 — b 2 

For values of [ m \ > -, there are two real tangents to a 
a 

hyperbola ; for \m\ < - the tangents are imaginary ; for 

m = ± - , there is only one tangent and its point of tangency 
a 

is at an infinite distance, or, as noted before, the lines y = ± - x 

a 

are asymptotes of the curve. 

The method of this article is employed in deriving tangents ; 

the equations given under A, B, and C above are used mainly 

in proving geometrical properties of these curves. Note that 

if these equations are used as formulas, they apply only to 

curves of the type given; y — mx -\ — gives the tangent only 

m 

to a parabola of the form y 2 = 4=ax (a may be positive or nega- 
tive). Similarly, y — mx ± -\/a 2 m 2 + b 2 gives the tangent of 
slope m only to the ellipse, 

- + ^ = 1, (a 2 may be less than b 2 ). 
a 2 b 2 v J J 



336 UNIFIED MATHEMATICS 



PROBLEMS 

Find tangents of slope 2 and of slope — 3 to the following 
three curves. Follow the method of article 4. 

1. x 2 + y 2 — 10 x = ; find the points of tangency. 

2. 3y 2 — 4ic — 6y = 0; find also the normal of slope — 1. 

3. x 2 -\-3y 2 — Ax — 6y = 0; find the diameter joining the 
two points of tangency. 

4. xy — 25 = 0. Find the tangents of slope — 2, and the 
points of tangency. Find the tangent of slope m. For what 
values of m are the tangents imaginary ? Plot 10 points on 
this curve. Where do all points of this curve lie ? Find the 
intersections of y — .01 x + 5 with this curve. 

5. Find the tangents at the extremities of the right focal 
chord of y 2 = 8 x ; where do they intersect ? Similarly in 
y 2 = 4 ax. Is this true of any parabola ? Explain. 

6. Find the tangents at the extremities of either right 
focal chord of ~ + ^- = 1 ; 

where do they intersect ? Similarly with 

x 2 , y 2 i , / b 2 ' 
— h — = 1, at [ ae, — 
a 2 b 2 ' \ a 

Where do these tangents intersect ? What change is necessary 
to prove this property for the hyperbola ? Explain. 

7. Find the perpendicular from the focus of y 2 = 8 x to the 
tangent of slope 2 ; where do they intersect ? Similarly for 
the tangent of slope m. State what you have found as a 
property of any parabola. 

x 2 v 2 

8. In the ellipse 1- ^- = 1, find the perpendicular from the 

focus to the tangent of slope 2 ; find the point of intersection ; 
note that it is a point on the circle, x 2 -f y 2 — 25. Prove that 



TANGENTS AND NORMALS 337 

the same is true of the perpendicular from the focus upon 
any tangent of slope m. Prove that in the ellipse 

a 2 b 2 
the perpendicular from the focus upon any tangent meets it 
on the major auxiliary circle. 

9. Follow the directions of problem 8 with the hyperbola, 

16 9 

and — = 1, 

a 2 b 2 ' 

making necessary changes. 

10. Find the angle between y=5x—7 and the parabola 
y 2 = Sx. 

The angle between a straight line and a curve is defined to 
be the angle between the straight line and a tangent to the 
curve at the point of intersection. 

Xote. — Solve for the points of intersection ; write the equation of 
tangent at each point of tangency ; find the angle between each tangent 
and the line y = 5x — 7. See article 8, chapter 15. Check by finding 
the slope angle of the lines y = 5 x — 1 and of the tangent lines. 

11. Find the angle between y = 2x — 5 and x 2 + y 2 = 100, 
at each point of intersection. Check as in problem 10. 

5. Tangent at a point (x h y^) on a curve given by an algebraic 
equation. — On any curve a line joining a point P x to a point 
P 2 is called a secant ; obviously this secant cuts the curve in 
two distinct points. If P 2 approaches P x along the curve, the 
secant changes, approaching more and more nearly, in general, 
a definite limiting line. The limiting position of the secant is 
called the tangent to the curve at P x . 

The analytical method of obtaining this limiting line is as 
follows : 

Take P x (x u y^ any point on the curve ; 



338 UNIFIED MATHEMATICS 

Take P 2 as {x x + h, yi + k) also on the curve ; the chord P 1 P 2 

k 
has the equation y — y x = - (a? — x x ). 

k 
Eind the value of - conditioned by the fact that P 2 and P 1 

both lie on the curve, by substituting (x l} y x ) and (x ± + h, y x -f- k) 
in the given equation and subtracting, member for mem- 
ber. 

k 

This value of - will be found, in general, to have a definite 
h 

limiting value as k and h approach zero ; this limiting value is 

the slope of the tangent. 

The method outlined applies to any curve given by an 

algebraic equation. 

y 2 — 4 ax ; Pi(x l7 y{) on curve ; P 2 (x } + h f y x + k) on curve ; 

k 
V — 2/i = 7 (p - x i), chord joining P X P 2 . 

(2/1 + W = 4 a(x 1 + h), or yf + 2 ky x -f k 2 = 4 ax 1 + 4 oTi, 
since P 2 is on curve. 

y* = 4 aa?i, since P 1 is on curve. 

2 A;?/! + A; 2 = 4 a/i, by subtraction. 

- = gives the slope of the chord joining P 1 to P 2 . 

h 2 y x + A; 

A: 
Let 7i approach 0, A; also approaches 0, but - always equals 

and this value approaches more and more nearly to 

2 y x + A; 

— — or — as a limit ; this limit is the slope of the tangent. 
2 2/1 yx 

2 a 

y — y 1 = — (x — Xj) is the tangent to 

2/i 
y 2 = 4 ax at (a^, i^) on curve. 



TANGENTS AND NORMALS 339 

This equation may be simplified, 
Dili - !Jl~ = 2 ax — 2 ax u 

y^y = 2 ax + y<f — 2 ax x ; but y-f = 4 ax, whence 
y x y = 2 a(x + a?i)j which is the tangent equation. 

By precisely this method, the tangent to 

^ + t = l ja ,t(x 1 ,y l ) 

a 2 o l 

on curve has been found, in section 11, chapter 18, to be 



a 2 b 2 



The tangent to 



x 2 u 2 -, , , N . x x x Ihll ^ 

— = 1, at (xi, Vi) on curve is — ^^ = 1. 

a 2 b 2 v J J a 2 b 2 

The tangent to 

Ax 2 + By 2 + 2 £# + 2 2fy + (7 = 0, at (a^, ?/i) on cury e is 
^a; + ^ + G(x + a?i) + F(y + 2/i) + C = 0. 

The tangent to 

Ax 2 + 2 #£?/ + By 2 + 2 #a; 4- 2 ify + (7 = 0, at (x^y^ on curve is 

Ax,x + Sfay + Vxx) + By lV + 0(a> + a^) + F(?/ + y x ) + C = 0. 

All the preceding are embraced in the last formula, as 
special cases. The final form should be remembered and used 
as a formula. 

The above special forms for the equations of the tangents to 
conies given by equations in standard form may be used to 
derive tangential properties of these curves. Some of these 
properties are touched upon in the problems below and will 
recur in the next chapter. 



340 



UNIFIED MATHEMATICS 



6. Tangential properties of the parabola. — 

?/ 2 = 4 ax, any parabola P x VPf 

y 1 y = 2 a(x-\-x 1 ), tangent at Pi(x 1} y x ) on curve, cutting the 
axis at T, the directrix at Q, the vertex tangent at S. 



y 



yi = — Ql (x — x^j normal P^N, cutting the axis at JSF. 

£ a 



llipfrffi i^^^WIH^tW 


IwM^fflw IW HI I It! 


<»-- --T7 ,'*/l& 7^=1 -^\ 


::::::::::::^::::::i^:z^::r:::::::r:::::::::::::::::::::::: 


::::::::::::::: ;£:_$ p r£:::::::M::::::::H:::::::::::::::::j^ 

::::::::::::::::::::^Z-::::::::::::::::::::::±:::::::::::::::: 


L 2 M 

:::::::::^::::g::::^::::::::::::::::::::::::::::::::::::::::: 


:::::::::::: ::::::::::^-::::::::::::::::::::::::;r::::::=:==: 


:::::::~:::::::::::::::::::::!:s;-::::::::::::::::::::::::::::::: 


eeHieHlieeiim^meNisHiiiHiiiiiiH 



Tangential properties of the parabola 

T and N are obtained as the ^-intercept of tangent and 
normal, respectively; 

T is (- %, 0) ; AT is (^ + 2 a, 0). 

Hence "RZ 7 = FJf, the tangent cuts off from the vertex on 
the axis the same distance that the perpendicular from the 
point of tangency on the axis cuts off from the vertex. This 
gives a simple method of drawing a tangent to a parabola : 
drop the perpendicular P X M to the axis ; extend the axis from 
the vertex, making VT = VM, the length of the intercept; 
P x T is the tangent. 

P 1 F=P 1 Z=x 1 + a. 
TF=x x + a = FN. 



TANGENTS AND NORMALS 341 

Hence Z FTP 1 = Z FP X T, base angles of an isosceles A. 

Now Z FTP 1 = Z TP X Z, alternate interior angles of 
parallel lines, etc. ZFP l T=Z TP L Z, i.e. the tangent 
bisects the angle between a focal chord and a line parallel to 
the axis ; the normal P X X bisects the supplementary angle 
FP X R, making Z FP X X= Z XP.R. 

Further S is the mid-point of TP X (since VS Y is parallel to 
P 1 31 and bisects the side TM). 

.-. FS is perpendicular to TP X ; the perpendicular from the 
focus to any tangent meets it on the yertex tangent. 

QF being drawn, A QFP X = A QZP X . 

Hence QFP X is a right angle. 

Extend P X F to cut the parabola at P 2 ; draw P 2 Z 2 to the 
directrix, and P 2 Q. 

A P 2 Z 2 Q = A P 2 FQ. 

Hence P 2 Q bisects the angle Z 2 P 2 F and is the tangent. 

Further Z P 2 QP X is a right angle, since it is half of the 
straight angle about Q. 

Summary of tangential properties of the parabola 

1. The tangent bisects the angle between the focal radius 
and a line parallel to the axis ; the normal bisects the in- 
scribed angle between the focal radius and a line parallel to 
the axis through the point of tangencj^. 

2. The perpendicular from the focus of any parabola on any 
tangent meets it on the vertex tangent. 

3. Tangents at the extremity of a focal chord meet on the 
directrix, and at right angles. 

4. The focal chord is perpendicular to the line joining the 
focus to the intersection on the directrix of the tangents at 
the extremities of the focal chord. 



342 



UNIFIED MATHEMATICS 



7. Tangential properties of the ellipse and hyperbola. — 

/y»2 o/2 /v»2 -3j2 

— (- £- = 1, any ellipse, £- = 1, any hyperbola. 

a 2 b 2 a 2 b 2 J J1 



^ + M = i ? 



a 2 6 2 ' 



tangent at (a? 1? ^) on curve, cutting the principal axis in the 
point T. 




Tangential properties of the hyperbola 

The tangent bisects the angle between focal radii to the point of tangency 



y - 2/i = + pg{x - ok) ; y - y x = -^(x - x 2 ), 
giving the normal at x lf y x on curve. 



Tis 



»i 



0); i^is {fx 1: 0). 



In the ellipse, 
a 2 

.1'! 



In the hyperbola, 



F X T = - - ae; F 1 N= ae - e 2 ^. F l T=ae - — ; F l N=e 2 x 1 - ae. 
F 2 T= ae + — ; F 2 N= ae + e 2 x v F 2 T= ae+ - • 2^= e^-r ae. 

#! X x 



TANGENTS AND NORMALS 343 

The lengths F y y and F 2 N are seen to be proportional to the 
Lengths P x F y and P y F 2 . 

P y F y F y N ■ a - ex t ae - e 2 x y 

1 J = — — , since = : 

P y F 2 F 2 N a + ex y ae + e\v y 

a? 

ae 

«. .1 i P x F y F,T ■ a-ex y x y 

bimilarlv l l = ^— , since -' == — 

1 P y F, F.,T a + exi a 2 . 

^ l — h ae 

If a line from the vertex of a triangle divides the opposite 
side into segments proportional to the adjacent sides, the line 
bisects the angle of the triangle ; hence the tangent and 
normal at any point on the ellipse and hyperbola bisect inter- 
nally and externally the angle between the two focal radii to 
the point. 

Another method of constructing the tangent is to construct 

a 2 

— . In the ellipse the major auxiliary circle x 2 -f- y 2 = a- is 

x i 

drawn and the tangent at P z {x y , y 2 ) on this circle x y x -f- y 2 y = a 2 

has the intercept — ; draw the tangent at (x y , y ) to the circle, 
x l 

cutting the X-axis at T '; connect T with P y on the ellipse. 

In the hyperbola a^ > a, so that this construction cannot be 

used ; from M(x y , 0) on the X-axis a tangent to the circle 

a 2 

x- + y 2 = a 2 intersects it at a point U whose abscissa is — . since 

x x 

OT = a 

a x\ 

Summary of tangential properties of the ellipse, hyperbola, and 
parabola, regarding the parabola as having a second focus at 
an infinite distance on its axis of symmetry. 

1. The tangent to an ellipse, hyperbola, or parabola bisects 
the angle between the focal radii to the point of tangency. 

2. The perpendicular from the focus upon any tangent 
meets it on the circle having the center of the conic as center, 



344 UNIFIED MATHEMATICS 

and passing through the principal vertices. In the parabola 
this circle has an infinite radius and so reduces to the tangent 
line at the vertex of the parabola. 

3. Tangents at the extremities of a focal chord meet on the 
directrix. 

4. The focal chord is perpendicular to the line joining the 
focus to the intersection on the directrix of tangents at the 
extremities of the given focal chord. 

PROBLEMS 

1. Find the tangent to the curve xy = 25 at the point (5, 5) 
by the method of article 5. 

2. Find the tangent to x 2 = Sy at the point (5, - 2 g 5 -) by the 
method of article 5. 

3. Find the tangents to the curves in problems 1-3 of the 
preceding set of problems at a point (x x , y x ) on each curve by 
the general formula. 

4. Find the tangent to the curve x 2 = 8 y at a point 
( x u 2/i) on the curve ; note that (a?!, yi) satisfies the equation 
of the given curve ; find a second equation which the point 
(a*!, 2/1) must satisfy if the tangent obtained is to pass through 
(5, 3) which is not on the curve ; solve the two equations as 
simultaneous and thus obtain the point of tangency of a tan- 
gent from (5, 3) to the given curve. 

5. Find tangent and normal to the curve x 2 — 10 x — 8 y — 5 = 0, 
at the point whose abscissa is 2 ; find the tangent of slope — 2 
to this curve. 

6. What tangential property of parabolic curves makes 
them useful in reflectors ? Explain. Prove the property. 

7. Write the equation of a hyperbola having the foci and 

x 2 v 2 
vertices of the ellipse — + ~ = 1 as vertices and foci, respec- 

tively ; find where these curves intersect ; write the equation 
of a tangent to each of the curves at one of the points of inter- 
section ; discuss these lines. 



TANGENTS AND NORMALS 345 

8. Write the equation of the tangent at a point (x l} y{) to 
each of the following curves ; use the general formula ; time 
yourself. 

a. 4:X 2 — 6x + 9y 2 + 5y = 0. 

b. ±x 2 -6x-9y 2 -5y = 0. 

c. 4 x 2 — 6 x — 5 y = 0. 

d. 4x 2 +6xy + 9y 2 —6x — 5y = 0. 



x- 



,2 



— (- ^-p = 1 ; it is not necessary to clear of fractions as 

•Jg- and 2V can be thought of as co-efficients of x 2 and y 2 . 
f. x 2 '-6x-4:y 2 -Sy + 7 = 0. 

9. Find the tangents to the first three curves in the pre- 
ceding exercise at the points where these curves cut the 
a>axis. 

10. Find one point on each of the curves of the eighth 
problem and write the equation of the tangent at that point. 




From Tyrrell, History of Bridge Engineering 
Elliptical arch bridge at Hyde Park on the Hudson 
The span is 75 feet and the rise is 14.7 feet. Note that the reflection 
completes the ellipse. 



CHAPTER XXII 

APPLICATIONS OF CONIC SECTIONS 

1. General. — Numerous applications of the conic sections, 
viz., circle, ellipse, parabola, and hyperbola, have been indi- 
cated in the problems given under the discussion of each curve. 
In general it is the tangential properties of the curves and 
the further geometrical peculiarities of these curves that make 
them so widely and so variously useful. The fact that simple 
geometrical properties are connected with curves given by 
algebraic equations of the first and second degrees in two 
variables seems to imply a certain harmony in the universe of 
algebra and geometry. 



2. Laws of the universe. — In 1529 the Polish astronomer- 
mathematician, Copernicus (1473-1543), rediscovered and 
restated the fact, known to ancient Greeks, that the sun is 
the center of the universe in which we live j he conceived the 

346 



APPLICATIONS OF CONIC SECTIONS 347 

planets to move about the sun in circular orbits. About a 
century later the great German astronomer, Kepler (1571- 
1630), was able to establish the following laws of the universe : 

1. The orbits of the planets are ellipses with the sun at one 
focus. 

2. Equal areas are swept out in equal times, by radii from 
the sun to the moving planet. 

3. The square of the time of revolution of any planet is 

proportional to the cube of its mean distance from the sun ; 

T 2 d 3 
i.e. — = — , if T x and T 2 are the periodic times of two planets, 

1 2 ^2 

and c?! and d 2 the diameters of their respective orbits. 

Kepler's work was made possible by that of all his pred- 
ecessors, particularly the Greek mathematicians who had so 
thoroughly discussed the properties of the conic sections, and 
further by the work of the Dane, Tycho Brahe (1546-1601), 
whose refined observations gave the necessary data. 

Newton (1642-1727) completed the work of systematizing 
the laws of motion in the universe in which we live, showing 
that the attraction of an}' two bodies for each other is in- 
versely proportional to the square of their distance apart and 
directly proportional to their masses. Newton showed further 
that this assumption leads to the elliptical motion in the case 
of the sun and any planet. 

The paths of comets which pass but once are known to be 
parabolas, or possibly hyperbolas with eccentricity close to 1. 

3. Projectiles. — The first approximation to the path of a 
projectile is a parabola. Indeed for low velocities, below 
1000 feet per second, the path is almost parabolic even with air 
resistance. The parametric equations of the path of a projec- 
tile shot horizontally with a velocity of 1000 feet per second, 
neglecting air resistance, are, in terms of t, the number of 
seconds of flight, as follows : 

x = 1000£, 

y = - 16 1\ 



348 



UNIFIED MATHEMATICS 



-400 



100 



■200 



-30( 



500 



600 



700 

I 



900 



1000 



s 



><; 



SSffiffiSOTffiSiffi 



When a projectile is shot at an angle a with the horizonal, 
we have shown that there is a horizontal component of veloc- 
^^^^^^^^^^^^^^.^^^ ity, v cos a, and 

ST ■ ■" 



a vertical compo- 
nent of velocity, 
vsina. The equa- 
tions of the path 
of this projectile 
shot from the 
ground as as-axis 
are as follows : 




Path of projectile shot horizontally from a tower 

1000 feet high; initial velocity of 50 feet per 

second 

X = 50 1, 
y = -16t\ 



x = v Q cos a ■ t, 
y = v sin a ■ t 

-16* 2 . 

Problem. — 
Find the path of a 
projectile thrown 
with a velocity 
of 50 feet per 
second horizon- 
tally from the 
top of a tower 
1000 feet high. 



constitute the parametric equations of the path, the axes being 
taken through the top of the tower. Giving to t values 
t= 1, 2, 3, ••• 8, these equations determine the position of the 
projectile after t seconds. 

t = 0,l, 2, 3, 4, 5, 6, 7, 8 determines the following points 
upon the parabola : 

(0, 0) (50, -16) (100, -64) (150, -144) (200, -256) 
(250, - 400) (300, - 576) (350, - 784) and (400, - 1024). 



APPLICATIONS OF CONIC SECTIONS 



349 



Since x = 50 1, y — — 16 t 2 , y 



-»■(*) 



for all values of t. 



x 2 = — 156.25 y is the equation of the parabola in standard 
form ; the coordinates (x h y x ) of any point obtained by sub- 
stituting a given value for t in the parametric equations above 

will satisfy this equation since t 2 —(— ) = — -*-. 

On any ordinary coordinate paper the curve x 2 = — 156.25 y 
can be plotted only as x 2 = — 156 ?/. 

The drawing shows very plainly that the projectile reaches 
the earth when t = 7.9 seconds, approximately ; solving 

-1000 = -16** (y = -16« 2 ,y = -1000), 
gives t 2 = 62.5, 

t = 7.91- 

The motion of a falling body is a special case of the equa- 
tions above, y = — 16 1 2 gives the space in feet covered in time 
t seconds by a freely falling body, falling from rest. 

4. Illustrative problem. — For a bullet shot at an angle of 
30° with a velocity of 1000 feet per second the equations are : 
x = 866 t, 
y = 500t-16t 2 . 

This bullet will, on a level plain, remain in the air until y = ; solving 
gives the value for the time of flight. The range is given by inserting the 
value of t so found to find x. 

The velocity of 1000 feet is equiva- 
lent to two separate velocities, one 
vertical of 500 feet per second, one 
horizontal of~ 866 feet per second. 
These are x and y components of the 
velocity. If no other force acted on 
this projectile, the path would be a 
straight line, given by 
x = 866 1, 
u = bQ0t. 




Vertical and horizontal compo- 
nents of a given velocity 



But since gravity acts, diminishing the vertical velocity, the total y is 
given by y = 600 * - 16.0», 



350 UNIFIED MATHEMATICS 

the — 16 i 2 being due to the effect of gravity. The fall in 1 second due 
to gravity is independent of the upward motion. 
The path is given by 

x = 866 t, 

y = 500t -16 V. 

When y = 0, the projectile is on the ground, the x-axis. 

t(500-16£)= 0, t = 0, or £ = — ; the first value, t = 0, means 

16 

simply that the projectile is shot from the ground, t = — = 31^, is the 

number of seconds the projectile is in the air. Finding x when t = 31^ 
gives the horizontal distance, or the range. Eliminating t gives the 
Cartesian form of the equation of the parabola 

x 

x = 866 t, or t = , whence 

866 

y = 500— - 16f— V, which reduces to 

866 \mo) 

(x - 13530) 2 = - 46870(?/ - 3906). 

The numerical work is somewhat tedious in such a problem, and it 
is indeed in most practical problems. The labor can be materially 
shortened by remembering that since the initial velocity is probably 
correct only to the second significant figure, correct here to hundreds 
of feet, and since y is taken as 32, instead of 32.2, an error of defect in 
the division of f of 1 °fo , the error by excess in the quotient will be also 
| of 1 °fo. 

PROBLEMS 

1. Of the planets Mercury is nearest to the sun. The mean 
distance of Mercury (=&) is 36 million miles; e = .2056 ; 
compute the equation of the orbit referred to the principal 
diameter as axis ; find the distance of the sun from the center 
of the path. 

2. Venus is the planet which is second in order of distance 
from the sun ; the mean distance is 67.27 million miles ; e = 
.0068, compute the equation and constants as in the preceding 
problem. 

3. Compute the orbit of Mars and focal distance; mean 
distance is 141.7 million miles ; e = .0933. 



APPLICATIONS OF CONIC SECTIONS 351 

4. Knowing that the earth has a time of revolution of 
365.256 (use 365.3) days and that its mean distance is 92.9 
million miles, compute by Kepler's third law the times of 
revolution of the planets in the three preceding problems. 

5. Discuss the maximum and minimum speed of the earth, 
assuming that the angular velocity is constant ; note that the 
focal distance is 1.5 million miles, and the variation will 
depend upon the different lengths of the radius. 

6. Assuming that the big gun which bombarded Paris had a 
range of 75 miles when pointed at an angle of 45°, find the 
initial velocity from the equations, 

x = .707 vt, 

y = .707 vt - 16.1 V. 
Insert in these equations x = To times 5280 and y = and 
solve for v and t ; the values obtained are the theoretical 
initial velocity in feet per second and the time of flight in sec- 
onds, neglecting air resistance. 

7. A body falls a distance of 10,000 feet ; find the time of 
fall. 

8. A body is thrown up vertically with a velocity of 100 
feet per second ; discuss the motion. 

5. Reflectors. — The fact that the tangent to a parabola 
bisects the angle between a focal radius and a line parallel to 
the axis leads to diverse uses of the parabola. Kays of light 
from the sun or from a star meet a parabolic mirrored surface 
whose axis is directed towards the sun or star in rays parallel 
to the axis of the parabolic surface ; these rays converge at the 
focus of the parabola and are by this means intensified. 

In an automobile reflector and in searchlights the conditions 
are reversed ; rays emanating from the central light at the 
focus are reflected in raya parallel to the axis. 

A ia\ of light directed towards one focus of a hyperbolic 
surface striking the surface is reflected towards the other 



352 



UNIFIED MATHEMATICS 



focus, since the tangent bisects the angle between the focal 
radii. This property is used by astronomers to re-focus the 
rays of light from the parabolic mirror at a point which does 
not lie between the parabolic mirror and the sun or star. An 
elliptical mirror beyond F l might be used for the same purpose. 




Parabolic and hyperbolic reflectors at the Detroit Observatory 

The curvature of the large parabolic mirror is greatly exaggerated on 

the diagram. 



The parabolic mirror here pictured, is in use at the Detroit 
Observatory, Ann Arbor ; the diameter of the mirror is 37.5 
inches ; the focal length is 19 feet ; the focal length of the 
hyperbolic mirror used is 5 feet ; the second focus of the hy- 
perbola is 2 feet behind the vertex of the parabola and at this 
point, F 2 , the rays are directed into a spectroscope. 

Sound rays are entirely similar to light rays so far as reflect- 
ing properties are concerned. In an auditorium it is desired 
that the sound waves should be thrown out from the reflecting 
walls about the stage in parallel lines to all parts of the build- 
ing; the reflecting surfaces have parabolic sections with the 
focus at the center of the stage. 

This is the case in the Hill Auditorium at Ann Arbor, 
Michigan ; axial sections of the hall made by planes are para- 
bolic in form, having the focus at the center of the stage. 



APPLICATIONS OF CONIC SECTIONS 



353 



6. Architectural uses of conies. — The intimate connection 
between beauty of form and numerical relations is undoubtedly 
illustrated by the " golden section." The most satisfactory 
dimensions of a rectangle from an artistic standpoint are such, 




, Hell Gate bridge, over the East River, New York City 

The largest parabolic arch in the world, in one of the most beautiful 
bridges of the world ; the arch has a span of 977.5 feet, height 220 feet. 



so it is accepted by those qualified to judge, that the longer 
dimension is, approximately, to the shorter as the shorter is to 
the difference between the two. In other words, if the width 
is given, the desired height is found by the " golden section," 
i.e. by dividing the line in extreme and mean ratio. Thus 
for width 40, the height x is found by solving the equation, 



x 



40 = _ 
x 40 — x 



this gives a quadratic equation for x. Note that if a square 
is cut off at one end of this rectangle a similar rectangle 
remains ; so also if the square on the longer side is added to the 
rectangle a larger rectangle similar to the original one is 
formed. 



354 



UNIFIED MATHEMATICS 



We have found a certain connection apparently existing be- 
tween simplicity of form and simplicity of algebraic equation. 
Thus the straight line is represented by the simplest algebraic 
equation in two variables, the first degree equation ; the circle 
which is the simplest curved line to construct is represented 
by a particularly simple type of quadratic equation ; to other 




The Williamsburg bridge over the East River, New York 

Longest suspension bridge in the world ; the parabolic arc of each 
18-inch cable is 1600 feet in span by 180 feet in depth, width 118 feet ; 
weight of the whole 1600-foot span is 8000 tons. Largest traffic of any 
bridge in the world. 



types of quadratic equations in two variables correspond only 
three further curves, viz., ellipse, parabola, and hyperbola. 
That these second-degree curves, the conic sections, are 
artistically satisfactory is evident from the extended use 
which has been made of these forms by artists, ancient and 
modern. 

In the construction of arches it is found that beauty of 
geometric form is intimately connected with simplicity of 



APPLICATIONS OF CONIC SECTIONS 355 

algebraic equation. The parabola and the ellipse have wide 
uses in construction not only because of beauty of form, but 
also because of purely mechanical adaptation to the stresses 
and strains caused by the weight of arch structures. A 
recognized authority 1 on bridge building, states that " arches 
must be perfect curves," and warns against the use of false 
ellipses. 

The fact that in many of the greatest bridges of the world 
the pure ellipse and parabola appear so frequently is an indica- 
tion of the wide acceptance of the theory that elliptical and 
parabolic arches are beautiful in form. The great Hell Gate 
Bridge of New York has for the main arch a true parabola 
(see problem 11, p. 317) ; London Bridge has five elliptical 
arches as the fundamental part of the sub-structure. Even 
the hyperbola has been used, but that only rarely. Let it be 
noted that partly because of the greater ease in design the 
circular arch is much more common, and the approximation 
to ellipse or parabola by using several circular arcs with dif- 
ferent centers is also common. 

No less than four distinct and different uses of the parabolic 
arc are found in the construction of bridges and trusses. The 
suspension bridge with a parabolic cable is one type ; the 
parabolic arch with vertex below the roadway of the bridge 
is a second use ; the parabolic arch intersecting the roadway 
is the third type ; and the parabolic arch entirely above the 
roadbed, as a truss, is a fourth type. 

Elliptical arches and less frequently parabolic are commonly 
used in the design of large foyers of theaters and in other 
large halls. 

Parabolic and pure elliptic arch forms are used, although 
not as frequently as circular and horseshoe forms, in the 
design of seAvers. Even complete perfect ellipses have been 
used (see problem 6, p. 356). 

1 Mr. fi. H. Tyrrell, of Evanston, Illinois, Artistic Bridge Design, 
Chicago, 1912. 



356 UNIFIED MATHEMATICS 

PROBLEMS 

1. Solve the quadratic in the preceding article and check 
by drawing a diagram. 

2. Find the width, x, of a rectangle whose height is 40, such 
that 40 is the " golden section " of the width, giving beauty of 
form of the rectangle. 

3. The Panther-Hollow Bridge, Pittsburg, has a parabolic 
arch, 360 feet in span with a rise of 45 feet= Draw the parab- 
ola. Assuming that the vertical chords are spaced every 
twenty feet and rise 15 feet above the vertex, find their 
lengths. 

4. In the preceding problem the smaller arches leading to 
the bridge itself are probably elliptical. The width of these 
arches is 28 feet and the height of the arch proper about 8 
feet. Draw these arches. 

5. A parabolic sewer arch used in Harrisburg, Pa. (designed 
by J. H. Fuertes) has dimensions of 6 feet in width by 4 
feet high. Construct ten points. 

6. A vertical elliptical sewer in Chicago, Western Avenue 
sewer, constructed 1910, has dimensions 12 x 14 feet. Draw 
the figure. 

7. Draw an elliptical and a parabolic arch, width 100 feet, 
height 30 feet ; compare. 

8. Draw to scale the parabolic arc of the Williamsburg sus- 
pension bridge, 1600 feet in span by 180 feet in depth. Find 
the equation in simplest form, choosing proper axes. Find 
the lengths of four vertical chords from cable to the tangent 
at the vertex of the arc. 

7. Elliptical gears. — On machines such as shapers, planers, 
punches, and the like the actual movement during the opera- 
tion of shaping, planing, or punching is desired to be slow and 



APPLICATIONS OF CONIC SECTIONS 



357 



steady, and the return motion is desired to be much more 
rapid. Circular gears give a uniform motion, but elliptical 
gears permit the combination of slow effective movement with 
quick return. The two ellipses are of the same size and are 




Two positions of elliptical gears, mounted on corresponding foci 

Note that the right-hand ellipse swings through a large angle, in this po- 
sition, as compared with the left-hand one. 

mounted at the corresponding foci. In every position then 
the two ellipses will be in contact since the sum of the focal 
distances in either ellipse equals 2 a, always, and this also 
equals the distance between the two fixed foci which are 
on the axes of rotation. 

PROBLEMS 

1. Draw three positions of two elliptical gears, each being 
an ellipse 6 inches by 10 inches. Determine maximum and 
minimum radii. When the ellipse is turned 5 times a minute, 
what is the fastest linear speed of a point on either ellipse ? 
Note that it is given by using as radius the maximum radius, 
FP in our figure. Find the slowest speed. 



358 



UNIFIED MATHEMATICS 



2. In the Sandwich hay-press of our illustration the diame- 
ters of the ellipse of the elliptical gears are 21 T \ inches and 




Elliptical gears on a hay press; the slow pressure stroke 



18^ inches ; plot the graph and discuss maximum and mini- 
mum linear speed, given that the angular velocity is twenty to 
twenty-two revolutions per minute. 







p 


~"1 


,^ 


.'"'■■:#'ff ■ -/ jP' ^ 



Elliptical gears on a hay press; quick return motion 



8. Applications in mechanics and physics. — The applications 
of the conies in mechanics and physics are very frequent. 
Thus the equation giving the period of a pendulum, 



-*4. 



I ,„ 7T 2 

- or t 2 = — 



• I 



APPLICATIONS OF CONIC SECTIONS 359 

(see page 317) is the equation of a parabola, when g is taken as 
constant. Similarly the velocity of water flowing from a tube 
or over a dam depends upon the height or head of water above 
the level of the tube or dam ; the relation is v 2 = 64.4 h, where 
h is measured in feet ; this also is a parabolic relation. 

The bending-moment at any given section of a beam sup- 
ported at both ends and uniformly loaded varies at different 
points on the beam, being greatest at the middle. These mo- 
ments are computed graphically in the case of a bridge, being 
given by a so-called parabola of moments. This parabola for 
a bridge of length I, uniformly loaded with a weight of w per 
foot, is given by the equation, 

M = -i- wl 2 — 1 iv • x 2 , 

wherein x is the distance from the center of the bridge. The 
parabola is plotted across the length I of the bridge, with the 
vertical ordinate at the mid-point representing the maximum 
moment. 

Thus if a bridge is 100 feet wide and uniformly loaded 
2 tons per foot, the moment at any point x distance from the 
center of the bridge is given by the formula 

M = " x U -1000a; 2 

8 

= JXl0 7 - 1000 a; 2 . 

Draw the corresponding parabola, choosing appropriate units. 

When a rotating wheel is stopped by the application of 
some force which reduces the velocity uniformly per second, 
the equations giving the number of revolutions before the 
wheel comes to rest correspond closely to the equations of 
motion of a body moving under the acceleration of gravity. 

6 represents numerically the angle covered in time t seconds, 
the body having an initial rotational speed of <d revolutions 
per second and the velocity being retarded every second by 
k revolutions per second. Here again we have an equation be- 



360 UNIFIED MATHEMATICS 

tween and t represented by a parabola. The time in which 
this body comes to rest is obtained by dividing the initial 
velocity by the uniform decrease in velocity per second,' i.e. 
by the acceleration (or retardation). 

The relation between pressure and volume of a perfect gas, 
temperature being constant, is given by the equation : 

p . v = k ; 

in words the volume is inversely proportional to the pressure. 
Plotting points gives points on a hyperbola of which the p-axis 
and the v-axis are the asymptotes. 

Such illustrations could be multiplied, but many relations 
of this character, e.g. the ellipsoid of inertia, require consider- 
able technical explanation which would go beyond the limits 
of this work. 

9. Quadratic function. — The graph of the quadratic function, 

ax 2 -f- bx + c, is the locus of the equation, 

y = ax 2 -f- bx 4- c. 

b V fb 2 - 4 ac y 
y = a * + — ] -a - 



=a[x + 



la) \ 4 
b V 5 2 -4ac 



4ac \ 
a 2 J 



2a 4a 



, b 2 - 4 ac ( , b 



Aac f 

= a[ i 

a V 

j4ac\ 
a / 



6 V V , h 2 - 



2a7 aV 4 

The graph of y = ax 2 + bx -+- c is a parabola ; ^ -f- — = is 
the axis. If a is positive the parabola opens up ; the vertex 

is V( — 77—, — | ; if ft 2 — 4 ac is negative, the vertex is 

V 2a 4a J 

above the jp-axis and no real value of x makes y = 0, since the 

graph does not cut the axis. If a is negative, the parabola opens 



APPLICATIONS OF CONIC SECTIONS 361 

down ; if b 2 — 4 ac is negative, the vertex is below the avaxis 
and again the graph does not cut the avaxis. If b 2 — 4 ac = 
the graph is tangent to the sc-axis. Evidently b 2 — 4 ac < 
is the condition that ax 2 + bx + c = should have imaginary 
roots ; b 2 — 4 ac — is the condition for equal roots ; and 
b 2 — 4 ac > is the condition for real roots. 

PROBLEMS 

Plot to the same axes the graphs of the functions in prob- 
lems 1, 2, and 3. Discuss. 

1. y = 5x 2 + 2x- 7. 

2. y = 5 x 2 + 2 x + 7. 

3. y = 5x 2 +2x-\-± 

4. If a wheel is rotating at the rate of 800 revolutions per 
second, and a force acting continuously reduces the speed each 
second by 40 revolutions per second, find the time in which it 
will stop, and the number of revolutions made during the 
retarded motion. Use the formula given in article 8. 

5. G-iven that 10 cubic centimeters of air are subjected to 
pressure, at a pressure of 1 atmosphere the volume is 10, 
hence pv = 10 is the equation connecting volume and pressure. 
Plot the graph of this for values of p from \ atmosphere to 5 
atmospheres' pressure. 

6. Plot the parabola of moments for the Hell Gate Bridge, 
assuming a uniform loading of 2 tons per foot. Do not 
reduce tons to pounds, but use ton-feet as units of moment. 
The equation is M= 490 2 — x 2 , taking 980 as the length of the 
bridge. 

7. Plot the parabola of moments of the Panther-Hollow 
Bridge in problem 3 of the preceding exercise, assuming 2 
tons per foot as loading. The equation is M = 180 2 — x 2 . 

8. Find the equation of a parabola whose focal length is 
19 feet. Draw the graph to appropriate scale. This is the 
parabola which, revolved about its axis, gives the parabolic 



362 UNIFIED MATHEMATICS 

reflector, previously mentioned, which is in use at the Detroit 
Observatory. (See page 352.) 

Find the equation of a hyperbola which has the same focus 
as this parabola, the axis of the parabola as transverse axis, and 
the second focus on the axis at a distance of 2 feet on the other 
side of the vertex. This is the hyperbola which, revolved 
about its axis, gives the hyperbolic mirror mentioned. 

The parabolic mirror has a diameter of 37 inches. Find the 

abscissa for the ordinate — — , thus finding the depth of the 
mirror. 

9. Plot the parabola y 1 = 70.02 x. This is the parabola 
which is fundamental ■ in the construction of the Hill Audito- 
rium. (See page 352.) The plane of the floor cuts the side 
walls in this curve ; so also the intersection of the ceiling and a 
plane passed vertically through the main aisle of the hall. 
In the plans the computations of ordinates for given abscissas 
are made to the thirty-second of an inch. Compute the focal 
ordinate. This is the radius of the circular 'arch over the 
stage. Compute the ordinates for x = 21, 26, 31, 51, and 71, 
and express, in feet and inches. 

10. The Italian amphitheaters are, in general, elliptical. 
The Colosseum in Rome (see illustration, page 288) is an 
ellipse with axes of 615 and 510 feet. Draw the graph to scale. 
On the same diagram and with the same center and axes of 
reference draw the arena, of which the dimensions are 281 
feet by 177 feet, to the same scale. Note that the minor axis 
of the arena is almost the " golden section" of the major axis, 
i.e. Ill is approximately a mean proportional between 281 
and 281 less 177. Find the mean and compare. 

11. The bridge at Hyde Park (see illustration, page 346) is 
elliptical, with a span of 75 feet and an arch height of 14.7 
feet. Draw this elliptical arch to scale. 



CHAPTER XXIII 

POLES, POLARS, AND DIAMETERS 

1. Definition. — The straight line 

Ax x x + B Vl y + G(x + a*) + F(y + Vl ) + C = 
is called the polar of P 1 (x lf y x ) with respect to the conic 

Ax* + Bf + 2 G?a; + 2Fy + = 0. 
The point P x (o5 1? ^x) is called the pole of the line. 

2. Fundamental property of polar lines. — If the polar of 
^i( x ii Vi) with respect to the given conic passes through 
F 2 (x 2 , y 2 ), then reciprocally the polar of P 2 (x 2 , y 2 ) will pass 
through PxOi, yi ). 

This fundamental property of polar lines enables one to 
prove complicated geometrical theorems for conies with a 
minimum of machinery. The proof of the theorem is itself 
simple, for substituting in the polar of P\{x^ y^, the co- 
ordinates (x 2 , y 2 ), we have that 

Ax x x 2 + By r y 2 + G(x 2 + a^) + F(y 2 + y{) + C = 0, 

if the polar of P 1 passes through P 2 . However the polar of 
P 2 is, by definition, 

Ax 2 x + By 2 y + G(x + x 2 ) + F(y + y 2 )+C=0, 

and substituting (a^, y x ) gives precisely the preceding expres- 
sion, which is of value ; hence P x (x 1} y{) is on the polar of 

F 2 (x 2 , y 2 ) • 

3. Geometric properties of the polar. — If Pi(x h y{) lies on the 
curve, the polar is the tangent at that point. (See the preced- 
ing chapter.) 

363 



364 



UNIFIED MATHEMATICS 



If P x (a^, ?/i) lies outside of the conic, the polar is the chord 
of contact of tangents from P x . 

Let P 2 (x 2 , y 2 ) be the point of tangency of a tangent drawn 
from P 1 ; 

by definition, the polar of P 2 is the tangent at P 2 (x 2 , y 2 ) ; 
by construction, the polar of P 2 passes through P 1 ; 




The polar of any point outside a conic is the chord of contact 

by the fundamental reciprocal property, since the polar of P 2 
passes through P 1} the polar of P 1 will pass through P 2 . 

Similarly, calling P 3 (x 3 , y 3 ) the other point of tangency, the 
polar of P x fa, yi) will pass through P 3 . 

Since the polar of P x is a straight line and since it passes 
through the two points of tangency, it is the chord of con- 
tact joining these two points. 

If Pj (x lf 2/1) lies inside, or outside, or on the conic, the polar 
is the locus of the intersection R(x\ y') of tangents drawn at 
the extremities of any chord passing through P lm 

Let R (a/, y') be the intersection of tangents at the extremities 
of any secant ; 



POLES, POLARS, AND DIAMETERS 365 

then, by the construction, the secant drawn is the chord of 
contact of B (x/, y') ; 

by construction, the polar of B (x f , y f ) passes through Pi(x u y x ) ; 

hence, by the fundamental reciprocal property, the polar of P x 
Avill pass through R (x', y') ; 

but the polar of P x is a straight line ; 

hence the locus of B (x r , y') is a straight line, the polar of 

A O'i, 2/i). 

By pure Euclidean geometry it is rather complicated to 
prove that the locus of the intersection of tangents at the ex- 
tremities of all chords of a circle passing through a fixed 
point is a straight line. The above proves this property for 
every conic. 

4. Diameter : definition and derivation. — The locus of the 
midpoints of a series of parallel chords in any conic is called 
a diameter of the conic. 

The method, applicable to any equation of the second degree, 
is given for a special case. Find the diameter bisecting chords 
of slope 3 in the ellipse 9x , +25 f = 900 _ 

Let y = 3 x -f- k 

represent any line of slope 3. Solve, as simultaneous, with 

9 x 2 -\-25y 2 = 900, which represents 
the conic. 

Substituting, 

234 x 2 + 150 kx + k 2 - 900 = 

is an equation whose roots are the abscissas, x l and x 2 , of the 
two points of intersection. 
Solving, 

x = ~ 15 ° k + V(150 k) 2 - 4(234)(fc 2 -~900) and 

_ _ 1.50 k - V(150 k) 2 - 4(234)(A: 2 - 900) 
468 



366 



UNIFIED MATHEMATICS 



For the midpoint (x' f y') of the chord, 
, _ x x + x^ _ —25 A; , 

the midpoint lies on the chord 
y = 3 x + k, 



hence, 



y 78 78 26 



The middle point of any chord, 

y = 3x + k, . 
is given by 

V — 25 1. 

X — j-gti, 

k 



y' = + — , and these two equations 




Any diameter of a parabola is parallel to the axis of the parabola 

constitute parametric equations of the locus of the midpoint. 
Eliminating k by solving for k and substituting (or by division 
here) , we see that for every value of k the 'Coordinates of the 
middle point satisfy the equation 



POLES, POLARS, AND DIAMETERS 



367 



Hence the middle point is on the straight line 

By precisely similar reasoning, the diameter bisecting chords 
of slope m in the conic, given by 

Ax 2 -f By 2 + 2 Ox + 2 Fy + C= 
is ^ic + m^ -f 6? + mF = 0. 

Applying this to the simplest standard forms of ellipse and 
hyperbola, we see that every diameter of a central conic passes 
through the center ; applying to the parabola, y 2 = 4 ax, we 
obtain my — 2 a = 0. This shows that the diameter of any 
parabola is parallel to the axis of the parabola, for when a 
parabola is rotated or moved to any other position any diame- 
ter moves with the curve, preserving its position relative to 
the axis of the curve. 



5. Reciprocal property of diameters in central conies. — In the 

conic 9 x 1 + 25 y 2 — 900 = above, the diameter bisecting 




Conjugate diameters ; each bisects all chords parallel to the other 

chords of slope 3 has the slope — ^. Now the diameter 
bisecting chords of slope — ^ has the slope 3, for by substi- 



368 UNIFIED MATHEMATICS 

tuition in the general formula, we have 9 x — -fa (25) y = 0, or 
y = 3x. 

Each of these diameters bisects chords parallel to the other. 
These are called conjugate diameters. In precisely similar 
manner this property can be established in any ellipse or 
hyperbola for the diameter bisecting chords of slope m. Do 
this for the diameter of the general conic above. 

PROBLEMS 

1. What is the equation of the chord of contact (polar) of 
(10, 0) with respect to 9 x 2 + 25 y 2 = 225 ? Solve this with the 
curve. This gives the points of tangency of tangents from 
(10, 0). Write the equation of the tangent at each of these 
points. This process illustrates an analytic method for finding 
a tangent from an external point to any conic, 

Ax 2 + By 2 + 2 Gx + 2 Fy + C = 0. Explain. 

2. Find the equations of the tangents to the following conies 
from the points given ; follow the method of problem 1 ; time 
yourself. 

a. y 2 -Sx = 0, from (-2, 6). 

5. y 2_Sx-6y-10 = 0, from (-3, 5). 

c. x 2 + y 2 -10x + 8y-59 = 0, from (18, 6). 

d. x 2 - 4 y 2 - 10 x + 8 y - 59 = 0, from (10, 5). 

e . X 2 + y i _ 25 = 0, from (1, 8) and from (2, 8). 

3. Prove in the ellipse — + *- •= 1 that if the diameter is 

1 a 2 b 2 

drawn through (xi, y x ), the tangents at the extremities of this 
diameter are parallel to the polar of Pi(jb 1 , yi). Call the 
points on the diameter (x 2) y 2 ) and (sc 3 , y s ), and note that they 
lie on the ellipse. Write the equations of the different lines 
mentioned. 

4. Prove the property mentioned in the preceding problem 

x 2 v 2 
for the hyperbola ^- = 1 and for the parabola y 2 — 4 ax = 0. 



POLES, POLARS, AND DIAMETERS 369 

5. In problem 2, write the equations of the diameters bisect- 
ing chords of slope 2 and of slope — i, using the general 
formula for diameter. 

6. In problem 5 write the equations of the conjugate di- 
ameters in the central conies. 

7. Draw the circle x 9 + y 2 — 36 = ; draw 5 secants 
through the point (4, 3) ; draw the tangents at the two inter- 
section points of each secant with the curve. The 5 points 
of intersection, one from each pair of tangents, should lie in 
a straight line. What theorem proves this ? 

8. Prove that the tangential parallelogram circumscribed 
at the ends of conjugate diameters of an ellipse b 2 x 2 + a 2 y 2 =a 2 b 2 
has a constant area. First show that the one end of the 
diameter conjugate to the diameter through Pi(x lf y r ) is 

P 2 ( — , — - ) ; find the equation of the tangent at P x (x lt ?/i) ; 

find the distance between (0, 0) and (x 1} y-^) ; find the equation 
of OPi ; find the perpendicular distance from P 2 to OP x ; by 
multiplication show that the area of this quarter of the given 
parallelogram is constant. 

9. Taking Pi(x ly y±) and P 2 (x 2 , y 2 ), which, by the preceding 

exercise, mav be written PJ ■—, — ], as the extremities of 

\ b a J 

a pair of conjugate diameters in the ellipse b 2 x 2 -\- a 2 y 2 = a 2 b 2 } 

show that the sum of the squares of OP x and OP 2 equals 

a 2 + b\ 

Hint. — Reduce the expressions for OPi 2 and OP 2 2 to common de- 
nominator, and use the fact that P\(xi, y\) is on the given ellipse. 

X 2 v 2 
10. In the hyperbola — — ^- = 1; the conjugate diameter to 

the diameter through Pi(x Y , y x ) on the hyperbola does not 
cut the curve itself. Prove this. 

The extremities of the conjugate diameter are taken as 
the points in which the conjugate diameter intersects the 



370 UNIFIED MATHEMATICS 

conjugate hyperbola *^ = — 1« With this definition the 

property of problem 8 can be proved to be true for the hyper- 
bola. State the method. What modification would you expect 
so far as the property of problem 9 is concerned ? 

11. Prove that the polars of all points on a diameter of any 
conic are parallel, comparing with problems 3 and 4 above. 

12. Show that tangents at the extremities of a series of 
parallel chords in any conic intersect on the corresponding 
diameter. 

13. Prove that any point on a diameter of the ellipse 

— I- — = 1 and the intersection of the polar of the point with 
a 2 6 2 f f 

the diameter divide the diameter length internally and ex- 
ternally in the same ratio. 

Hint. — Take (xi, y x ) and (— Xi, — y\) on the curve as the extremities 
of the diameter ; take the point of the diameter as the, point ( — ^ — - 1 , 

^ ^ x ) which divides the diameter externally in the ratio r ; find the 
l-r ) 

intersection point of the polar of this point with y = — x, and note that it 

Xi 

is the same as the point which divides the line joining (xi, y{) to ( — Xi, — y{) 
internally in the ratio r. The property holds for any conic. 

If through any point a secant to a conic is drawn, the point 
and the intersection of the polar of the point with the secant 
divide the chord of the conic formed by the secant internally 
and externally in the same ratio. The proof is somewhat 
more complicated than that of the preceding special case. 

14. Show that the tangential parallelogram to any central 
conic formed by the tangents at the extremities of a pair of 
conjugate diameters has its sides bisected by the points of 
tangency. An ellipse can be rather neatly inscribed in any 
parallelogram by drawing the ellipse tangent to the sides of 
the parallelogram at the midpoints. 



CHAPTER XXIV 

ALGEBRAIC TRANSFORMATIONS AND 
SUBSTITUTIONS 

1. Transformation of coordinates. — For varied reasons it is 
sometimes found desirable to change the location of the 
coordinate axes with respect to a curve which is given by an 
equation involving variables. Usually this shifting of the axes 
is for the purpose of simplifying the discussion of the geo- 
metrical properties of the curve in question. Thus the ellipse 

has been given in the form ^ '— + ^ *- = 1, but the 

a 2 5 2 

geometrical properties of the same curve are discussed with 

reference to the center (h, k) as origin, giving the equation 

a 2 ft* 

The axes may be subjected to a translation, giving new axes 
O'X' and 0' Y' parallel to the old axes ; or the axes may be 
turned through an angle a, giving new axes OX' and OY' 
about the old origin ; the two motions can be combined, execut- 
ing first the translation, usually, and then the rotation ; it is 
possible also to shift to new axes inclined at an oblique angle 
to each other, but the formulas involved are too complicated 
for an elementary work. 

2. Translation of axes. — Suppose the ic-axis fixed and the 
?/-axis moved parallel to itself to a new origin 0' at distance 
00' = h, from 0. Take P(x, y) as the coordinates of any 
point with reference to the original axes. Evidently, as the 
#-axis is unchanged, the y of this and every other point remains 

371 



372 



UNIFIED MATHEMATICS 



the same. Let M be the feot of the perpendicular from P to 
the a>axis ; then by our fundamental property of the distances 
between three points on a directed line 

OM= 00' + O'M. 





y->- -y 










i- (rr-O/) 


KlWl 


















<-* h -Q ~r' ~^ 


4-4- 


± ± 


- _ ± -± __ _ _ 



But OM 



o.a = k, 



Translation of axes 



the distance either posi- 
tive or negative 00'; 
while 0'M= x' by defini- 
tion. Hence, whatever 
the position of P(x, y), 
we have, 

x = x' + h. 
Similarly, if the a>axis is shifted parallel to itself by an 
amount Jc, 

y = y' + Jc. 
The two equations 

x — x' -\- h, 

y = y' + k, 

transform any equation given with respect to any axes, to a 
set of parallel axes having the point (h, h) as origin. 



3. Algebraic substitution in functions of one variable. 
Theorem. — Substitution of x'-\- h for x in any algebraic equa- 



tion of type a x K -+■ a x x r 



••• a n _ x x + a H = 0, n an integer, gives 



a new equation whose roots are 7i less than the roots of the old. 

The proof of this theorem depends directly upon the pre- 
ceding article. The substitution x — x' -f- h moves the y-axis 
h units, reducing the abscissas of all points by h if h is positive 
and increasing them by — h if h is negative. 

Illustration. — If the graph of y = x 3 — 2x 2 — 18 x + 24 is plotted, 
the substitution y = y and x = x' + 4 simply shifts the ?/-axis 4 units to 
the right, thus decreasing the numerical value of each root by 4. 
The new equation is 
y = (x' + 4)3 - 2(x + 4)2 - 18(x' + 4) + 24 = x' 3 + 10 x' 2 + 14 x' - 10. 



TRANSFORMATIONS AND SUBSTITUTIONS 373 

Now whatever number substituted for x makes x 3 — 2 x' 2 — 18 x + 24 = 0, 
it is evident that. 4 less substituted for x' will make 

(*' + 4) 3 - 2(x' + 4)2- 18(x' + 4) + 24 = 0. 

Graphically, of course, as we have indicated, the y-axis has been pushed 
4 units towards the right, and the abscissa of each point of intersection 
of the curve with the x-axis has been reduced by 4. 



INI 

i ■ . ' 


m 


1L 


1 1 i I U 1 1 1 ■ V 

' : -( 2,t-44-V-KT ■ 




i , . 


H- r- 

tt 


yp ' N^~ -4pfi — q: 

44-14- --BS 

L_____^^ — - W 


, 1 

mriTihMffH 


Jr T~ 


- ■ ' 


f-N-i- -- --- -W,- -- 

^42 L ^1-:e :i:^:\2::i :3 

§^^#=# 

JJ_M ____ 20 sZl 

__ET 


:::::::::::::::::::::£: 

lf|l 




- - 






_j 3: 














LL — 1_ 


1 II 1 1 1 


liMI iiiiiiMi-4e l|l||MMI1 





Graph of y - x 3 - 2 x 2 - 18 x + 24 

Similarly, in trie general equation above, when x' -f- A is sub- 
stituted for x, whatever number a satisfies the original equa- 
tion in x, a — h will satisfy the new equation in x'. 

Substitution of x' 4- h for x in any algebraic equation forms 
a new equation in x' whose roots are h less than the roots of the 
given equation. 

This type of substitution is used to facilitate the computa- 
tion of roots of numerical algebraic equations. 

A simple method of constructing the new equation in nu- 
merical equations will be explained below, in section 11 of the 
next chapter. 



374 UNIFIED MATHEMATICS 

PROBLEMS 

1. Show that the formulas of transformation given trans- 
form the equations of ellipse and hyperbola having (Ji, k) as 
center to the simpler form without first-degree terms. 

2. Transform the equation (y — 3) 2 = S(x + 2) to the point 
(3, — 2) as new origin, new axes parallel to the old. 

3. By translation of axes transform the equation 

into a new equation in which the first-degree terms are lacking. 

4. Find the equation of the line Sy — 4 a; -f- 6 = 0, referred 
to parallel axes through the point (3, 2). 

5. Compare the slope of a straight line referred to new 
axes by translation, with the slope referred to the old axes. 
Compare intercepts. Compare the slope of a tangent at a 
fixed point on any curve with respect to new and with respect 
to old axes. 

6. Given the expression for the volume of 1000 cu. cm. of 
mercury at 0° C. when heated to t° C, v = 1000 + .0018 t (see 
page 63), transform to parallel axes with the point (t = 0, 
v = 1000) as new origin ; find the new equation in v' and t. 
Does v' represent volume ? 

13 1 

7. Given v = 1054 -f — — , the velocity in feet per second of 

sound in air at t° centigrade, transform to parallel axes with 
(32°, 1054) as new origin ; discuss the equation. 

8. The equation jc 3 - 2 x 2 - 18 x + 24 = (page 373) was 
found to have a root between x = 4 and x = 5 ; transform 

y = x^ - 2 x 2 - 18 x + 24 = 
to parallel axes through (0, 4) and the new equation in x' will 
have a root between and 1. Compute this root to tenths by 
substitution. 

9. The equation x 3 — 2 x 1 — 18 x + 24 = has a further 
root between 1 and 2. Compute this root to one decimal place 
by the process explained in the preceding problem. 



TRANSFORMATIONS AND SUBSTITUTIONS 375 

li . Find the roots of 2 # 3 + 6 x 2 — 10 x — 8 = 0, as in prob- 
lems 8 and 9 by considering the graph of the equation 
y = 2 X s -f- 6 x 2 — 10 x — 8, when referred to new axes. (See 
problem 4, page 99.) 

11. Transform the following equations to parallel axes, 
having (h, k) as the new origin ; determine (h, k) so that the 
terms of the first degree shall disappear. 

a. 5x 2 + ±xy-y 2 -8x-5y-10=0. 

b. 5 x 2 + 4 xy + y 2 — 8 x - 5 y — 10 = 0. 

c. #?/ — 7 a; — 10 ?/ — 5 = 0. 

& 4:x 2 -6x- y 2 -Sy~10 = 0. 

12. Transform the following equations to parallel axes 
naving (h, k) as origin. Can you determine (h, k) so that the 
terms of the first degree shall disappear ? Why not ? (See 
problem 13.) 

a. £x 2 -6x-8y-10 = 0. 

b. 4 x 2 + 4 xy + y 2 — 8 a> — 5 ?/ - 10 = 0. 

13. Show that if an equation of the second degree contains 
no first-degree terms, the origin is the center of the curve by 
showing that if (x u y^) is any point on the curve (— x 1} — y x ) 
is also on the curve. 

4. Rotation of axes. — The formulas for sin (a + (3) and 
cos (a -f (3) give very neatly the relations which exist between 
the coordinates (x, y) of a point referred to the old axes and 
the coordinates (x f , y') referred to the new axes. Take P(x, y) 
any point referred to the original axes ; let a be the angle of 
rotation through which the axes are turned ; let ft be the angle 
which the line OP makes with the x' or new x -axis. By 
section 4, chapter 15, for all values of a and (3, 

cos (« + /?)= cos a cos ft — sin a sin (3, 

OP cos (a + fi)= OP cos a cos (3 — OP sin a sin /3. 



376 UNIFIED MATHEMATICS 

But OP cos (a + /3) = x) OP cos j3 = x' ; OP sin (3 = 2/' ; 

hence, . , . 

a? = x cos a — y sm «. 

Further, sin (« + /?) = sin a cos /? + cos a sin /?. 

Multiplying by OP, and substituting, 

y = x' sin a +y' cos a. 

These same relations might also have been obtained by pro- 
jection ; they hold for every position of the point P. 

x = x' cos a — y' sin a, 
y = x f sin a + y' cos a, 

effects the rotation through the angle a, and refers any equation 
in two variables to new axes inclined at an angle a to the old 
axes. 

5. Every second-degree equation in two variables represents a 
conic section. Proof. — To prove this theorem we need only to 
show that the equation 

(1) Ax 2 + 2 Hxy + By 2 + 2 Gx + 2 Fy + C = 

can, by rotation of axes, be transformed to an equation of the 

type (2) A& + By* + 2 Gx + 2 ify + C = 0. 

Every equation of this latter type represents, as we have 

shown, either circle, ellipse, parabola, or hyperbola or seme 

limiting form of one of these curves. 

Substituting, , , . 

x = x' cos a — y' sm a, 

y = x' sin a + y' cos a, 

the equation Ax* -f 2 Hxy + By 2 + 2 Gx + 2 Fy + C = 
becomes 

J.(a/ cos a — y' sin a) 2 +2 i?(V cos a — y' sin a)(V sin a+ y' cos «) 
-f- B{x' sin a+ y' cos a) 2 + 2 6r(#' cos a. — y' sin a) 
+ 2 F(x' sin a + y' cos a) + (7 = 0. 



TRANSFORMATIONS AND SUBSTITUTIONS 377 

Collecting terms, we have, 

(A cos 2 a -f- B sin 2 a + 2 H cos a sin a)x' 2 
-f- (^4 sin 2 « + 5 cos 2 a — 2 H cos a sin a)/ 2 
+ [— 2 .4 cos a sin « -f 2 i7(cos 2 a — sin 2 a) + 2 5 cos a sin a]x'y f 
+ (2G cos « + 2 Fsin a)x' +(2 i^cos a - 2 # sin <%' + C = 0. 

Let JV*. + 2 £T'a;y + By* + 2 # V + 2 jPy + C = 

represent this equation. 

We wish to show that it is always possible to find an angle 
a for which H' becomes 0. 

2 TT 
Setting H' = 0, leads to the equation tan 2 a = , 

noting that cos 2 a — sin 2 a = cos 2 a and 2 sin a cos ce = sin 2 a. 
Since iT, A, and 5 are real numbers and since the tangent of 
an angle can have any value from negative to positive in- 
finity, it follows that there is always some angle 2 a for which 

2 TJ 

tan 2 a = — . There are in fact always two positive angles, 

less than 360°, 2 a and 2 a + 180°, which satisfy the given re- 
lationship. By turning through a, or a + 90°, one half of 
either of these angles, the equation Ax 2 -\-2 Hxy + By 2 -{- • •• = 0, 
reduces to an equation of the type A'x' 2 -f B'y' 2 -f- ••• = 0, 
with the coefficient of the x'y' term equal to 0. The angle 
a of turning can always be selected as a positive acute 
angle, since if tan 2 a is positive, 2 a may be taken as an acute 
angle, and if tan 2 a is negative, 2 a may be taken as an obtuse 
angle of which the half-angle a will be acute. 

Illustrative problem. — What angle of rotation will remove 
the xy term from 3 x 2 + 6 xy — 5 y 2 = 100 ? 

2H 6 



tan 2a = 



A- B 8 



cos 2 a = \ ; cos a = v$ (1 + cos 2 a) = — — - , 

VlO 



since — y/\{\ — cos 2 a) = — — < 
v'10 



378 



UNIFIED MATHEMATICS 



We select a acute, as noted, hence the positive values of the radical are 
^aken. The formulas of transformation become, 



x = — — x' 
VlO 
1 



(x' + 3 y<). 
VlO VlO VlO 

Substituting, we have, 

&(3x' - y'Y + j%(Sx> - y')(x> + Sy')- &(a>' + Sy') 2 = 100. 



VlO 
VlO VlO 



_1 

VlO 

1 

vro 




The hyperbola 3 x 2 + 6 xy - 5 y 2 = 100, or 4 x' 2 - 6 y " 2 = 100 

In combining terms, do not write the expansion but preferably combine 
like terms by inspection. 

Here the coefficient of x' 2 is fir + io — to 5 of x 'v' tne coefficient is 
— it + it ~~ f $, ,or 0, which checks ; for t/' 2 we have T 3 7 — i|. — ^. Our 
equation becomes, 

4x' 2 -6?/ 2 = ioo, 



x' 2 y' 2 
25 16.67 



1. 



This curve is plotted with reference to the new axes, inclined at an angle 
ct, tana = J, to the x-axis. The coordinates of a point (x', y') on this 
curve, considered with respect to the new axes, satisfy the new equation 



TRANSFORMATIONS AND SUBSTITUTIONS 379 

— — -K = 1 ; when considered with reference to the old axes as (x, y) , 

25 16.(37 

the coordinates satisfy the original equation. Thus the coordinates of the 

intersection with the original x-axis (5.8, 0) satisfy the original equation ; 

this point with reference to the new axes has the coordinates 

5.8x3 
x' = x cos a + y sm a = — =- , 

VlO 

— 5.8 

y' = ~ x sm a + y cos a = — — , 

VlO 
or (5.5, — 1.8). The values for (&', y') in terms of (x, y) can be con- 
ceived as obtained by rotating through the angle — a. 

PROBLEMS 

1. Find the equation of the curve xy — Ix + 3y — 15 =0 
when referred to axes making an angle of 45° with the given 

axes. Note that a is 45° : sin a = cos a = — - ; rationalize 

V2 

denominators after substituting. Plot the new axes at the 
angle indicated and plot the graph of the new equation, obtained 
by substitution, with reference to the new axes. 

2. Find the equation of the curve 

9x 2 + 24;xy + 16y 2 -6x-15y = 
with reference to axes making an angle arctan -| with the old 
axes. Note that sin « = f and cos a = f ; in substituting take 
the fraction I as a factor in the value of both x and y and, 
after substituting, combine terms by inspection without writ- 
ing each expansion separately. 

3. Find the equation of the curve 

59 x* - 24 xy + 66 y 2 + 72 x - 396 y + 444 = 0, 
when referred to new axes such that the new a>axis makes an 
angle whose tangent is f with the old axis of abscissas. 

4. In the equation 4;xy — 8x -\-10y + 7 = make the gen- 
eral substitutions which effect the turning of the axes through 
an angle a, and determine a so that the coefficient of the x'y f 
term shall disappear. 



380 UNIFIED MATHEMATICS 

6. Nature of the conic Ax 2 +2 Hxy -f By 2 + 2 Gx + 2 ity + C= 0. 

A central conic is one which, has a point which is such that 
every chord passing through this point is bisected. If this 
point be taken as origin of coordinates, it follows that if (x', y') 
is on the curve ( — x', — y') is also on the curve. A substitu- 
tion, x — x'-\-h and y = y' +- k, which causes the terms of the 
first degree in our equation of the second degree to disappear 
gives the equation Ax' 2 -+ 2 Hx'y' +- By' 2 +- C = 0. Now what- 
ever point (V, y') satisfies this equation (— x', — y') will also 
satisfy the equation, and hence the new origin is the center of 
this conic. 

The substitution x = x' +- h and y = y' +- k gives two linear 
expressions in h and k as coefficients of the new x' term and y' 
term, and these are set equal to zero and solved for h and k to 
determine the center. 

2^+2^ + 2^=0 
and . 2Hh + 2Bk + 2F = Q 

are the two equations which determine the center. 

If the two equations which serve to locate the center repre- 
sent two parallel lines in h and k, the conic has no center and 

A J-f 

is a parabola. This condition is that — — = — -, or that 

H 2 -AB = 0. When H 2 -AB=0, the terms Ax 2 +- 2 Hxy + By 2 
form the square of a linear expression in x and y. 

Further it is shown below that if H 2 — AB < 0, the conic is 
an ellipse, and \i H 2 — AB > 0, the conic is a hyperbola. The 
conditions determining the nature of the general conic are as 

follows : TT0 at) ^ n ^^^ 

H 2 — AB < 0, ellipse, 
H 2 — AB = 0, parabola, 
II 2 - AB > 0, hyperbola. 

These are the conditions that there should be no points on 
the curve at infinity, one point at infinity, and two directions 
giving infinite points. They may be derived by substituting 



TRANSFORMATIONS AND SUBSTITUTIONS 381 

y = mx + k and determining values of m for which the quad- 
ratic has infinite roots ; it follows that for these values of m 
the line y = mx -f Jc will meet the curve in points infinitely 
distant. For the ellipse the values of m will be imaginary, 
and H 2 — AB < ; for the parabola the two values of m will 
coincide, and H 2 — AB = ; for the hyperbola the two values 
of m will be real and different, representing the slopes of the 
two asymptotes, and H 2 — AB > 0. 

A second and independent proof is given in the next article. 
There it is shown that the product AB' is positive when 
H 2 — AB is negative ; but when A! and B' are of the same 
sign the product is positive and the curve in x' 2 and y' 2 , not 
involving x'y', is an ellipse. Similarly the product A'B' is 
negative when H 2 — AB is positive, and the curve represented 
by the transformed equation is a hyperbola. 

7. Central conies ; abbreviated process of transformation. — 

Substitution method. — Determine the center ; transform to 
parallel axes with the center as new origin ; determine a and 
substitute ; plot with reference to the final axes. 

Abbreviated method. — Determine the center (h, k) ; trans- 
form to (h, k) as new origin ; determine A' and B' by solving 
as simultaneous the equations, 

A' + B' = A + B, 
- A'B' = H 2 ~ AB; 

select the pair of values of A' and B' such that A'— B' will have 
the same sign as H '; plot the new equation with reference to 
new axes having the origin at the center determined and the 
axes inclined at an angle a with the old axes, a being such that 

tan2«= 2JI . 
A- B 

Derivation of A' + B' = A + B ; - A'B' = H 2 - AB. 

A! = A cos 2 a + B sin 2 a + 2 H cos a sin a. 
B' — A sin 2 a + B cos 2 a — 2 H cos a sin a. 

By addition, A' + B' = A + B. 



382 UNIFIED MATHEMATICS 

The proof that — A'B' = H 2 — AB is somewhat long but 
not difficult. To the product — A'B' add 

H' 2 = [2 H (cos 2 a - sin 2 a) - 2 (A - B) sin a cos a] 2 , 

which does not alter the value since H ' is taken to equal 0. 

The expressions will combine to H 2 — AB. The student would 

do well to verify at least one of the coefficients. 

Since a is chosen as a positive acute angle, A' — B' has the 

same sign as H. for A' — B' 

= (A -B) cos 2 a + 2 # sin 2 a 



V 2^ 



cos 2 a -+- sin 2 a 



Now sin 2 a is positive, and cos 2 a has the same sign as 

2 jj a B • • • 

and hence the product of cos 2 a by is positive ; 



A-B x " 2H 

hence A' — B' is the product of 2 i? by the sum of two positive 
quantities and so is positive if H is positive and negative if H 
is negative. 

The equations. A' + B' = A + B 

-A'B' = H 2 - AB 

enable us to determine A' and B' by solving these as simulta- 
neous equations. Two solutions are found, and the solution 
is selected which makes A' — B' have the same sign as H. 

Only the new constant term, when transforming to (h, k) as 
new origin, offers any extended computation. This constant 
term 

Ah 2 + 2 Hhk + Bk 2 + 2 Gh + 2 Fk + G 

may be written 

h(Ah + Ilk + G)+k(Mi + Bk + F) + Gh + *% + C, 

which reduces to Gh + i^A; + C, since the other two expressions 
within parentheses were set equal to zero to determine the 
center. 

Illustrative pro! lem. — Find center, axes, and plot the conic, 

3 x* -h 6 xy + 5 y 2 - 12 x - 18 y - 24 = 0. 



TRANSFORMATIONS AND SUBSTITUTIONS 383 

Substituting (x' + h, y' + k) and selecting the coefficients of x' and y\ 
to set equal to zero, 

6 A + 6 fc - 12 = 0, 



6 h + 10 k - 18 = 0. 



Solving, k = f , fca + f 




The ellipse 3 x 2 + 6 xy + 5 y 2 - 12 x - 18 y - 24 = 

or 3x' 2 + 6x'y' + 5y' 2 - — = 0, or 7.16 x" 2 + .84 y" 2 = 40.5 
2 



C . the new constant, Gh + Fk + C a= - 6 • \ — 9 • f — 24 =— ^. The 

new (x', 2/') equation is 

3 x' 2 + 6 xy + 5 </' 2 - — = 0. Note that tan 2 a = — = - 3. 
2 — 2 

^1' + £' = 8, 
-.!'£' = 9- 15 =-6. 
Solving by substitution, 

-4'(8-4') =-6, 
^'2 _ 8 A > + 6 = o ; ^/ = 4 ± vTo. 
fi' = 4T VTO. 

A' — B' has the same sign as H ; hence the upper algebraic signs are 
taken, A' = 7.16, B> = .84. 



384 UNIFIED MATHEMATICS 

Our final equation is 

7.16z"2+.84 2/"2 = 40.5 

X !,2 yU2 = i 

5.66 48.3 
x" 2 . y'" 



= 1. 



(2.38)2 (6.95)2 

Some computation is unavoidable, and, in general, in practical applica- 
tions the results are rarely expressible in small and convenient integers. 

PROBLEMS 

1. Find the center, axes, and plot the conic, 

.5 x 2 - 6 xy + 3 y 2 + 12 x — 6 y - 30 = 0. 

2. Plot the following conies by turning the axes through 

2 77 

an angle a, tan 2 a = — , so as to eliminate the xy term, 

j± — Jo 

and thus obtain an equation to plot which can be put in stand- 
ard form. 

a. 4 x 2 + 4 xy + y 2 — 6 x -f- 8 y — 12 = 0. 

&. x 2 — 4 xy + y 2 + 2x-10 y - 11 = 0. 

c. 41 x 2 + 2±xy + 34:y 2 -26x-32y- 171 = 0. 

d. 4,xy—.3y 2 -7x— 10 y — 15 = 0. 

3. Apply the abbreviated method explained in section 6 to 
the central conies in the preceding problem; compare the 
numerical work involved by the two methods. 

4. Find five points on the first and second conies in prob- 
lem 2 by giving values to x and computing the corresponding 
values of y. 

5. Find the intercepts with the coordinate axes of each of 
the conies in problem 2 and verify your graphical construction 
by these points. 

6. In each of the conies of problem 2 find the points of 
intersection with the line y — mx + b ; determine the values 
of m for which one of the points of intersection should be at 
an infinite distance. In the case of the hyperbolas real 



TRANSFORMATIONS AND SUBSTITUTIONS 385 

values of m will be found ; substitute in turn each of these 
values for m and determine for what value of b the second 
point of intersection will move off to an infinite distance. 
This determines the two asymptotes. Explain. 

7. Apply the abbreviated method to the discussion of the 
following central conies, having the origin as center : 

a. x 2 + 2 xy + 4 y 2 = 16. 

b. 4x 2 -6xy-3y 2 = 10. 

c. 2x 2 — 4,xy-y 2 = —9. 

d. 5x 2 -3xy + y 2 = 24:. 

8. In the hyperbolas of problem 2, use the results of prob- 
lem 6 to show that the directions of the asymptotes are given 
by the factors of the terms of the second degree. 

8. The hyperbola as related to its asymptotes. — The equation 
of the hyperbola in simplest form, 

a 2 b 2 ' 
may also be written, 

(bx — ay) (bx -f- ay) = a 2 b 2 , 



whence. 



bx — ay bx -f- ay a 2 b 2 



Va 2 + 6 2 Va 2 + & 2 « 2 + & 2 



Since bx — ay = and bx -f- ay — represent the asymptotes 
of this hyperbola, the final form states that the product of the 
perpendicular distances of any point on the hyperbola from 
the two asymptotes is constant. The converse proposition is 
also true, viz., if a point moves so that the product of its 
distances from two intersecting lines is a constant, the point 
moves on a hyperbola of which the two lines are the asymp- 
totes. The proof of the converse is simply that the bisectors 
of the angles between the two given lines could be selected as 
axes of coordinates and, in consequence, the two lines would 
have as equations, expressions of the form y — mx = and 
y -\- mx = 0. Any point which moves so that the product of its 



386 UNIFIED MATHEMATICS 

distances from these two lines is a constant would satisfy the 
nation y _ mx y + mx = 



Vl + m 2 Vl + 



m- 



but this equation represents a hyperbola, and consequently 
the given locus is a hyperbola. 

It follows from the above argument that the equation of 
any hyperbola differs by a constant from the product of the 
first-degree expressions which, put equal to zero, represent its 
asymptotes. The terms of the second degree in the hyperbola 
can be factored always into real linear factors in x and y (not 
necessarily rational so far as the coefficients are concerned) 
which as lines have the slopes of the asymptotes. (See problem 
8 of the preceding list, and compare article 6.) A particularly 
simple type of hyperbola equation occurs quite frequently in 
practical problems and this type will be taken to illustrate the 
method which is, however, general. 

Illustrative example. — Plot the curve 

1 - x 

This equation may be written 

2/(1 —x) = x. 

The only term of the second degree is xy. Placing the factors equal to 
zero, we have x = and y = 0. The asymptotes are parallel to our co- 
ordinate axes. The equation can be written in the form 

(x - h) (y -k) = c. 

By inspection we note that the equation may be written 

0/ + l)(x-l) = -l. 

The asymptotes are given by y -f- 1 = and x — 1 = 0. 

The intersection point is the center of the given curve ; further points 
should be plotted by substitution of values in the original equation. 

This equation in i and d, i = — , represents the relation between a 

1 — d 

given rate of discount for any interval and the corresponding rate of in- 
terest. If a bank in lending money takes out interest in advance, giving 



TRANSFORMATIONS AND SUBSTITUTIONS 387 

to the individual not the face of the loan but that amount less the interest 
upon that amount for the given interval for which the note is to run, the 
bank is said to discount the note. The rate of interest which the indi- 
vidual pays is obviously greater than the rate d which is used as the dis- 
count rate ; the relation is 

1 -d 

In plotting the graph of this curve you would be interested only in values 
of i and d between .01 and . 10, and you would confine your attention to 
the first quadrant, taking £ inch to represent .01 on both axes. 

PROBLEMS 

1. Plot the curve p«v = 1000; show that it represents a 
hyperbola having the axes as asymptotes. This equation rep- 
resents the relation between the pressure and volume of a 
quantity of gas which at a pressure of 1 atmosphere has a 
volume of 1000 cubic units, the temperature being kept 
constant. 

2. Discuss the nature of the following curves, without 
making any transformation of axes ; in the hyperbolas give 
the slopes of the asymptotes, and in the parabolas the slope of 
the axis. 

a. 4 x 2 — y 2 — 8 y = 0. 

6. 4 x 2 - 8 y -10 = 0. 

c. 4z 2 -4x?/-?/ 2 -l00 = 0. 

d. 4z 2 — 4xy -\-y 2 = 100 y. 

e. 4 x 2 — 4 xy -f- if = 100. 
/. 4 x 2 - 4 xy + 4 y 2 = 100. 
g. 4x 2 -±xy-10x = 25. 

h. 4 xy — 7 x -f 10 y — 5 = 0. 

i. xy = 15. 

j. 4 x 2 + 4 y 2 = 81. 

7c. 3 x 2 - 12 x - 2 y 2 - 10 y - 15 = 0. 

I 3x 2 -12x + 2y 2 -10y-15 = Q. 



388 UNIFIED MATHEMATICS 

3. Transform to new axes so as to simplify the following 
equations to plot ; select the appropriate method of substi- 
tution adapted to each equation. 

a. x 2 + 12 xy 4- 4 if — 4 x - 24 y - 10 = 0. 

b. x 2 + 3 xy - 3 y 2 - 10 x - 15 y + 24 = 0. 

c. 4 x 2 - 4 xy + 7 ?/ 2 - 10 y + 4 a; - 25 = 0. 

d. 4x 2 -12a# + 9*/ 2 -6#-10 = 0. 

c. 2 x 2 + xy + 2 1/ 2 - 6 x + 6 y — 15 = 0. 

/. x 2 + 4 an/ - 2 ?/ 2 - 8 x + 20 y - 30 = 0. 

#. a? 2 — 3 xy — 7 a? = 0. 

h. 2x 2 -6x-\-5if-20y-10 = 0. 

4. Given that an aeroplane covers a distance of one hun- 
dred miles in t hours, its velocity in miles per hour is , 

i 

i.e. v = ; given that on different occasions the aeroplane 

covers 100 miles in 48 minutes (.8 hours), l^hour, 1 hour 6 
minutes, 1-J- hours, 1 hour and 24 minutes, 100 minutes, and 
2 hours, respectively, find the velocities and plot a curve giving 
the relation between v and t. Choose units so that you can 
read from the curve between the extreme values the velocity 
within 2 miles per hour when the time of flight for 100 miles 
is given. Note that the curve is a hyperbola. 

5. Given that an aeroplane covers on one trial 100 miles in 
48 minutes, on another trial 125 miles in 61 minutes, and 156 
miles in 71 minutes on a third trial, how could you compare 
graphically the corresponding velocities ? 

6. The air in an organ pipe vibrates in a manner some- 
what similar to the motion of a pendulum ; the number of 
such vibrations of the air in one second depends upon the 
length of the pipe and upon the velocity of sound in air ; the 

formula w = ^-, v in feet per sec. and I in feet, gives quits 



TRANSFORMATIONS AND SUBSTITUTIONS 389 



closely the number of vibrations. Plot the curve n = j , for 

values of I from 1 to 20 feet, choosing appropriate units. The 
curve gives the corresponding number of vibrations for pipes 
of different lengths. (See section 3, chapter 26.) 

7. Discuss fully the equation 

x 2 — 2 xy + y- — 10 y = 0. 

8. The curve of transition on a railroad track in passing 
from one straight track to another is sometimes taken as para- 
bolic, because of the fact that the slope changes uniformly 







— \i~ — 






W* — l 






-UM- 




























7* 








i — 




















/ 




II 


























-f- 1 


H-j- 




-j — ' — 










































































-stH- 











^p= — ^ 
















-X" 


' 








<*- 


















^^ 










f^~ 
















-7^- 










1 ~^— 

-^ 




— 










\ 1 


-^ 










s»* — 












































^ — a 










r^- 1 — 









~H — r 








**- 




\hS 






»* 








- 


~H — 


1/ 








V ^ 
^ 


3F 












— 




— 1 i 

— —4- -| 


— ^ 




-l— 












m 


X* 






— +— ^ 


7" 



















; i 






— i^— 










1 










— 






.yf" , 




, 


1 1 1 














1 i 1 



Parabolic transition curve on a railroad track 

The parabolic arc is used for vertical as well as for horizontal 
transition curves. 

with uniform increases of the horizontal length taken parallel 
to the tangent at the vertex of the parabola. Assuming that 
the track AV changes its direction by 60° to VB and that the 
transition points A and B from the straight line to the 
parabolas are taken on each track 500 feet from the point of 
intersection of the two directions, find the equation of the 
parabola. Note that the axis FFis inclined at an angle of 
120° to the extension of J.F; note that E, the vertex of the 



390 UNIFIED MATHEMATICS 

parabola, is midway between V and the point where the chord 
AB cuts the axis, since the tangent to a parabola cuts off from 
the vertex on the axis a distance equal to the distance cut off 
from the vertex on the axis by the perpendicular to the axis 
from the point of tangency. Find the equation of the curve 
with respect to the axis of the parabola as ?/-axis and the line 
through V at an angle of 30° with AV as sc-axis ; then 
transform to A V as a/-axis and a perpendicular to AV at V 
as i/'-axis by turning through an angle of — 30°, using the 
fundamental formulas for rotation of axes. 

9. Assuming that a railroad track changes its direction by 
40°, 30°, 20°, and 10° respectively, find the equations of the 
parabolic transition curves with transition points (A and B, as 
in figure) 500 feet from the intersection point of the two 
straight tracks. 

10. In going over a hill the form of curve to which the 
track bed is rounded is often made parabolic. When the grade 
is the same on both sides of the highest pointj the problem is 
precisely that of finding a parabolic arch. Assuming that in 
a horizontal distance of 5000 feet the hill rises 100 feet, find 
the equation of the parabola having the vertex at the highest 
point and passing through the point 100 feet lower at a hori- 
zontal distance of 5000 feet ; find the four intermediate ordi- 
nates at distances 1000 feet apart. 

11. An iron wire of diameter .2 cm. and length I cm., 
subjected to a tension T caused by a weight W grams, when 
caused to vibrate through its whole length has the number of 
vibrations determined by the equation 



- 1 /' 



W \ T=9S0W. 



.077 7T 

When the weight is fixed and the length is variable, this gives 
a hyperbolic relation between n and I. For W= 500 grams the 

equation is approximately n = . Plot and discuss. 



TRANSFORMATIONS AND SUBSTITUTIONS 391 

12. In the preceding problem suppose that I is fixed at 100 
centimeters and that W varies between 100 grams and 2000 
grams. What is the type of relationship ? What would be 
the curve obtained by plotting to to- and vi-axes ? 

13. The deck of any large vessel slopes from both bow and 
stern downwards towards amidships. The vertical section of 
the deck from bow to stern consists of two parabolas, having a 
common vertex at the middle of the ship. Plot the parabolas 
which are used for a vessel 400 feet long, having the highest 
point at the bow 8 feet above the vertex, and at the stern the 
deck 1 feet above the vertex. Use a different scale for y than 
for x, — at least twice as large. 

14. Name the following curves, giving such facts as you 
can by inspection : 



a. 

b. 
c. 
d. 

e. 


3 x + 2 y - 5 = 0. 

3 X 2 + 2 y - 5 = 0. 
3 x* + 2 y 2 - 5 = 0. 
3.r 2 - 5aj = 0. 
3 x 2 — o xy — 5 x = 10. 


k. 

1. 
m. 
n. 

0. 


(3a:+2y-5)(aj-3) = 

3 x 2 + 3 y 2 = 0. 

xy — 7 x + 6 y — 18 = 

p(3 — v) = 6. 

(l + 0(l-d) = l. 


= 10. 

= 0, 


f. 
9> 
h. 


3 x 2 + 3 y 2 - 25. 
3x*-6xy+3y 2 -ox=0. 
3 x°—6 xy + 3 y 2 - o = 0. 


p. 


1,1 . 
x y 




L 


3x 2 + 2y 2 + o=0. 
(3x+2y—5)(x-3)=0. 


v. 


F =331.7 A /l + 2 -|. 





15. The highway over the Michigan Central R.R. tracks 
and over the Huron River, on the Whitmore Lake road near 
Ann Arbor, is rounded off (in profile) to a parabolic arc, rising 
2.40 feet in a span of 240 feet. Show that the grade leading 
up to the arc should be a 4 % grade. Draw the arc to scale. 



CHAPTER XXV 

SOLUTION OF NUMERICAL ALGEBRAIC EQUATIONS 

1. Solutions of algebraic equations. — By a solution of an 
equation of the type a x n + a x x n_1 + a 2 x n ~ 2 -f- — a n _ x x + a n = 0, 
wherein n is a positive integer and a , a b a 2 ••• are constants, 
we understand a value which, substituted for x, reduces the left- 
hand member to zero. That such a solution always exists is 
proved by methods of higher mathematics. The theorem that 
every such rational integral algebraic equation has a root is 
called the fundamental theorem of algebra; it was first proved 
about a century ago by Gauss. The solution may be a real 
number or a complex number, and any constant coefficient may 
be real or complex ; the latter involves the square root of a 
negative quantity and so is not representable as the abscissa 
of any point on our axis of positive and negative real 
numbers. 

Certain types of algebraic equations are solvable in terms 
of the general constants which enter as coefficients. Thus 
ax + b = is solvable in terms of a and b, and ax 2 + bx + c = 
is solvable in terms of a, b, and c. It has been shown that the 
general cubic in one variable and the general biquadratic, or 
fourth degree equation, are solvable in this way, but the 
general equations of higher degree than the fourth are not 
solvable in this sense. 

The approximate numerical solution of the real roots of 
rational integral equations with numerical coefficients is readily 
obtained and we have indicated in Chapter II and again in the 
preceding chapter, section 3, problems 8-10, the general 
method by which such solutions are obtained by substitution. 

392 



NUMERICAL ALGEBRAIC EQUATIONS 



393 



Simplifications for purposes of computation will be explained 
in this chapter. 

2. Continuity. — The height of an individual is a continuous 
function of the age of the individual ; by this we mean that 
in passing from one height to another the individual passes 
through every intermediate height. A graph representing age 
as abscissas and heights as ordinates will be a continuously 



*& 



:S 



Four continuous graphs. One discontinuous 

Continuity in passing from a positive to a negative value. 

connected curve. Upon this curve corresponding to any 
selected age, a lt a period of time, there will be one and only 
one corresponding height, 7i 1} and corresponding to any second 
age, a 2 > a second ordinate, h 2 , representing height. The curve 
joining the two points (a b h^) to (a 2 , h 2 ) will be continuous and 
every intermediate height between the two given will be 
found to be represented by the ordinate corresponding to some 
age intermediate between the two given ages. 
The rational integral function of x, 

a x n + c^x 71-1 + ... 4- a n _iX + a n , 

in which n is any positive integer, is continuous between 
any two values of x, and will be represented by a continuous 
curve. This has been assumed in drawing the graph of 
y = x 3 — 2 x 2 — 18 x-\- 24, and in other graphs. The proof in- 
volves discussion somewhat too detailed and mathematically 
refined for an elementary course. 

The symbol f(x) will be used throughout the remainder of 
this chapter to represent a rational integral function of x of 
the type mentioned above. 



394 UNIFIED MATHEMATICS 

3. Graph of y = f(x) by location of points. 

Give to x the appropriate values, find the corresponding values 
of y, and plot the points, connecting by a smooth curve. (See 
pages 70-71.) 

Apply the remainder theorem, and employ synthetic 
division to determine values of the function corresponding to 
given values of x. 

4. Remainder theorem and synthetic division. (See page 25.) 

When f{x) is divided by x — <x, the remainder obtained by con- 
tinuing the division until the remainder does not contain x is 
equal to the original expression with a put for x. 

To divide fix) by x — a, employing synthetic division, 

a. Arrange fix) in descending powers of x and write the 
coefficients horizontally, including zero coefficients for missing 
powers below the highest power which occurs. 

b. Write + a under x — a, the divisor, placed at the left. 
Under the coefficients of fix) as written leave space for a 
second horizontal row and draw a horizontal line. 

c. Under the coefficient of the highest power of x, below 
the horizontal line drawn, place this coefficient again. Mul- 
tiply by -f a and add to the following coefficient to the right. 
Place the sum below the line, vertically under the second 
coefficient; use this number below the line as multiplier of 
+ a, and add the product to the third coefficient and continue 
this process until you have, placed numbers under every coeffi- 
cient (and the constant term) of the upper row. The final 
number which appears is the remainder and should be cut off 
by a vertical separator ; the numbers under the horizontal line 
are coefficients in order from left to right of the quotient 
when fix) is divided by x — a. 

Throughout this discussion a may be either positive or 
negative. 



NUMERICAL ALGEBRAIC EQUATIONS 395 

Illustrative Problem. — Divide x 3 — 2 x 2 — 18 x + 24 by x — 3, and 
use the remainder theorem to determine the value of this function of x 
when x = 3. x3 _ 2 x 2 - 18 x + 24 

x - 3) 1 - 2 - 18 +24 
+ 3) +3 +3-45 
T+l -15 (-21 
X 2 + x — 15 is the quotient and — 21 is remainder; — 21 is the value of 
x 3 — 2 x 2 — 18 x + 24 when 3 is substituted for x. Since 

x 3 -2x 2 -18x + 24=(x 2 + x- 15)(x-3)-21, 
we have, substituting 3, 

33 _ 2 . 3 2 - 18 • 3 + 24 ee (3 2 + 3 - 15) (3 - 3)- 21 
= 0-21. 

PROBLEMS 

1. Locate ten points upon the graph of y =2 a* 3 +3 x 2 — 9 x— 7. 
Take the ten points between x = — 4 and x = + 4, including 

J and — |- ; use the synthetic division method of finding the 
value of y except for x = 0, x = 1, and x = — 1. Plot the 
points and draw a smooth curve connecting them ; choose the 
y scale so as to keep the points on the paper. Locate the zeros 
of the function on the graph. 

2. Plot the graph of the function 2 x 3 4- 3 x 2 — 7 between — 3 
and +3. 

3. Plot the graph of the function 2 x 3 — 9 x — 7. Note 
where the graph crosses the axis of x, thus locating the roots 
of the equation 2 a,* 3 — 9# — 7 = 0. Factor 2 x 3 — 9 x — 7, 
dividing by the factor corresponding to the rational root which 
you have found; solve the resulting quadratic, and compare 
with the values found by the graph. 

4. Plot the graph of the function tf —2 x 3 +3 x 2 -18 x +21 ; 
select the appropriate interval to give the points of intersection 
with the a'-axis. 

5. Plot the graph of y = x 4 — 3 x 2 — 21 ; locate the zeros of 
the function on the graph. Solve as an equation in quadratic 
form x 4 — 3 x 2 — 21 = and compare the solutions obtained 
with the roots located graphically. 



396 UNIFIED MATHEMATICS 

6. Plot the graph of the function 4x 3 .— 3 x 4- .5 in the in- 
terval from — 1 to + 1 ; substitute for x the values — 1, — .8, 
-.5, -.3, -.1, 0, .1, .2, .3, .4, .5, .6, .7, .8, .9, and 1, finding 
the values in general, by the division method applying the re- 
mainder theorem. The roots of this equation represent the 
values of sin 10°, sin 50°, and sin — 70°. (See section 10, below.) 

7. Locate one root between and .1 of the equation 

4 ^-3^ + .05234=0, 
by substituting for x the values 0, .01, .02, .03, up to .1. The 
value .05234 is the sine of three degrees which we obtained in 
problem 5, page 245. One root of this equation gives the sine 
of 1°. 

5. Number of roots. — A value of a u for which/ (a x ) = 0, is a 
root of f(x) = 0. The remainder theorem applies and conse- 
quently (x — aj) is a factor of f(x) since the remainder when 
f(x) is divided by x — a x will be zero. Nothing in our argu- 
ment requires that a x be a real number. Hence, dividing f(x) by 
(x — a x ), a new equation of degree one less will be obtained. 
This equation, by the fundamental theorem of algebra, also has 
a root, a 2 , giving a quotient of degree n — 1. The number of 
such factors corresponds to the degree of the equation, n. 

Every rational integral equation of the nth degree has n roots, 
and no more. For no further value of x could make the product, 
a(x — ai)(x — a 2 )(x — a 3 ) ••• (x — a n ), equal zero without making 
one of the factors zero and thus coinciding with one of the roots 
given. 

6. Graphical location of real roots. — Any real root of a rational 
integral function of x equated to zero is a value of x which 
makes the ordinate in y =f(x) equal to zero. The points in 
which the graph of the function of x crosses, or touches, the 
#-axis correspond to real roots of the equation, f(x) = 0, or 
zeros of the function. 

Our assumption of continuity enables us to formulate the 
following theorem : 



NUMERICAL ALGEBRAIC EQUATIONS 397 

Between any two values x = a and x = b, for which the two 
corresponding values of f(x) are opposite in sign, there lies at 
least one real root of the equation f(x) = 0. 



m^ y I ■■ I I :: I : I 1 



m 



^ 



^ 



3 



g 



1 . 1 . 1....1. , 1 1 : \ ■ : l ..:;l;ii!l,ijl l i ! !Tr! - 



Four graphs passing continuously from y = - to y = ; one graph with a 

2 3 

discontinuity 

Thus to change continuously from + \ to — a, or from any positive 
value to any negative value, the function must pass through all values in- 
termediate, including 0. At this point where the function of x is 0, the 
graph of y = f(x) crosses the axis. 

Illustrative problem. — Locate the roots of 

XZ-2X* - 18a;+24 = 0. 

Plot the graph of y = x 3 — 2 x 2 — 18 x -f 24 by location of points. Give 
to x values from — 5 to + 5, find the corresponding values of ?/, and plot 
the points, connecting by a smooth curve. (See page 71.) Between x=l 
and x = 2, f(x) changes from -f 5 to — 12 ; there is a root between x = 1 
and x = 2 ; between x = 4 and x = 5 there is a root, as/ (4) is — 16 and 
/(5) is + 9 ; at x = — 4 there is a root, as/(— 4) is 0. 



7. Slope of y = f(x). — The (h, k) method of finding the 
tangent at a point (x ly y r ) on a curve applies, as we have 
stated in Chapter 18, section 11, to the graph of a rational in- 
tegral function of x. 

Thus in y = r 3 — 2x 2 — 18 # + 24, let (x ly y^ be any point 
on the curve and (x r + h, ^4- k) a second point. It is desired 
to find the slope of the graph at (x u y x ) 

^==^-2:^-18^+24, 

and y, + k={x l + hf - 2(x t + hf - 18(^ + ^ + 24, since (x h y x ) 



398 UNIFIED MATHEMATICS 

and (a?! + h, y l -\- k) are on the curve. Subtracting the upper 
from the lower equation, member for member, we have, 

k = h(3 xf - 4 x x - 18) + W(3 x x - 2) + 7i 3 , 

*? = 3 xf - 4 X! - 18 + h(S x 1 - 2) + /* 2 . 
7i 

Let h approach zero; the terms on the right containing h and 
h 2 will also approach zero, as the coefficients are constants. 

Limit - = 3 x x 2 - 4 x x - 18, as h = 0. 
h 

When x t = 1, the slope of the curve is — 19 ; when x x = 3, 
the slope is — 3 ; when x x = 4, the slope is + 14. 

A double root of any equation corresponds to a point at 
which the function is zero and the slope of the curve, ob- 
tained by the (h, k) method, is zero. 

8. Slope and maximum and minimum points. — When the 
slope is zero, the curve is for the instant parallel to the avaxis. 
This is a necessary condition for a maximum or minimum 
point, i.e. a point at which the value of the function attains a 
greatest or a least value in some interval which includes the 
point. 

This may be accepted by the student as graphically evident. 
A formal proof depends on the methods of the calculus, and 
rests essentially on the method used in finding the slope. 

PROBLEMS 

See the preceding list of problems. 

1. Find the slope at any point (a? l3 y x ) of each of the follow- 
ing curves and locate the maximum and minimum points 
on the curve by setting the slope equal to and solving for x x : 

a. y = 2x z + 3x 2 — 9x-7. d. y = 2 x 3 — 9 x - 7. 

b. y = 2x? + 3x 2 -7. e. y = 4 aj 3 - 3x + .5. 

c. 2/ = a^-3x2-21. /. 2/ = 4a; 3 -3a; + .05234. 



NUMERICAL ALGEBRAIC EQUATIONS 399 

2. Find the slope at any point (x 1} y^) on 

y = x* - 2 x 3 + 3 x 2 - 18 x + 21. 

This gives the slope m as m = 4 x^ — 6 a^ 2 -f- 6 x x — 18. Plot 
the graph of y = 4 x 3 — 6 x 2 + 6 a; — 18, and note that the zeros 
of this function are the values of x L for which the slope of the 
curve y = x 4 — 2 x 3 + 3 x 2 — 18 x + 21 is 0. These are values 
of x for which the original function has maximum and 
minimum values. 

9. Historical note. — The solution early in the sixteenth cen- 
tury of the cubic and biquadratic was the undisputed achieve- 
ment of a group of Italian mathematicians. Fiori, Tartaglia, 
and Cardan were involved in the solution of the cubic, while 
Ferrari, pupil of Cardan, solved the quartic. Not until the 
beginning of the nineteenth century was it shown that the 
general equations of higher degree are not solvable, this being 
the work of a brilliant young Norwegian named Abel. 

10. The cubic applied to angle trisection. — By higher mathe- 
matics it has been demonstrated that geometrical problems 
which can be solved by ruler and compass correspond alge- 
braically to problems whose solution can be effected by linear 
and quadratic equations and equations reducible to quadratics, 
i.e. by equations of which the roots will involve only quad- 
ratic irrationalities (square roots, and square roots of expres- 
sions involving only rational quantities and square roots). 
The trisection of an angle is a type of geometrical problem 
whose solution cannot be effected with ruler and compass ; it 
is possible to reduce the trisection of an angle to an algebraical 
problem, the solution of the cubic equation. 

Let the given' angle which is to be trisected be denoted, for 
convenience, by 3 a. Since this angle is given, the value of its 
sine is known. If the angle is given by a geometrical drawing, 
the ratio of the perpendicular h dropped from a point at a 
distance /* from the vertex on one side to the second side to r, 



400 UNIFIED MATHEMATICS 

i.e. - , gives the sine of the angle. Let the value of the sine of 
r 

the given angle be k. 

Given sin 3 a = k, find sin a. 
sin 3 a = sin (2 a + a) = sin 2 a cos a + cos 2 a sin « 

= 2 sin a cos 2 a + (cos 2 a — sin 2 a) sin a 
sin 3 a = 3 sin a — 4 sin 3 a. 
ft = 3 sin a — 4 sin 3 a. 
This equation is a cubic in the unknown sin a ; for convenience 
it may be written k = 3 x — 4 x 3 , substituting sc for sin a. 

There are, in fact, three solutions of the cubic and these 
three solutions correspond to the fact that k is the sine not 
only of 3 a, but also of 180° -3 a, and n ■ 360° + 3 a, and 
(2 n + 1)180° -3 a. 

Thus the cubic which would give the sine of 10°, trisecting 
the angle of 30°, is .5 = 3x - 4a^, or 4ic 3 - 3 x + .5 = 0. The 
same cubic would be obtained if it were desired to trisect the 
angle of 150°, or of 390°, or 750°, •••. There are an infinite 
number of angles which have this same sine, .5, but there will' 
be only three different values involved when the sine of the 
third part of each of these angles is found. In the equation 
4 x* — 3 x + .5 = 0, the roots represent sin 10°, sin 50°, and 
sin 250°. (See problem 6, page 396.) 

11. Closer approximation to located roots. — The method will 
be shown by a numerical illustration. 
The equation 

(1) a 3 -2a 2 - 18^ + 24 = 0, 

of which the graph is given on page 71, evidently has a root 
between 4 and 5. To form the new equation whose roots are 
4 less than the given equation, substitute a?' .+ 4 for x, giving 

(2) (x' + 4) 3 - 2(flj' + 4) 2 - 18(aj'+ 4) + 24 = 0. 
Assume that this gives 

(3) x' 3 + Bx' 2 + Cx f + D=0, in which B, C, and I) can be 
obtained by expanding and combining terms in (2). The left- 



NUMERICAL ALGEBRAIC EQUATIONS 401 

hand members of equation (3) and equation (2) are then identi- 
cal. Evidently, if x — 4 is substituted for x f in (2), it will give 
the original equation, and consequently, if x — 4 is substituted 
in (3), it will give the original equation. Substituting, we have 

(4) {x - 4) 3 + B(.v - 4) 2 + C{x - 4) +- 1) = 0, which is identi- 
cal with the original equation. 

If the left-hand member of this equation, i.e. the original, 
is divided by x — 4, the remainder is D and the quotient is 
(x — 4) 2 + B(x — 4) + C ; if this quotient is divided by (x — 4), 
the remainder is C ; if the quotient of the preceding division 
is divided by to — 4, the remainder is B. 
a_4) X s - 2a?-18x + 24 

+ 4 ) ^^ — + 8-40 The continued division by 

D 

x — 4 is effected by the 

synthetic process, explained 

in section 4, above. 



+ 


4 


+ '8 - 


-40 


1 + 
+ 


2 
4 


-10 (- 
+ 24 


-16 


1 + 

4- 


6 
4 


(+14 





c 



1(+10 B 

(5) x' 3 + 10 x' 2 + 14 a/ — 16 = is then the equation whose 
roots are 4 less than the roots of the original equation. This 
should be verified by substitution and expansion. The 
original equation has a root between 4 and 5. Hence this 
equation has a root between and 1. By trial of tenths, .1, 
.2, — .9, this equation is found to have a root between .7 and 
.8. Hence the original equation has a root between 4.7 and 4.8. 

Form the new equation whose roots are .7 less than the 
roots of (5). 

x > _ .7)1 +10 +14 - 16 (.7 
+ ,7 ) + .7 + 7.49 + 15.043 
1 +10.7 +21.49(- .957 

+ .7 + 7.98 
1 +11.4(+ 29.47 

+ .7 
1(+12.1 

(6) 2 3 + 12.1 z 2 + 29.47 z - .957 = 0. 



402 



UNIFIED MATHEMATICS 



Equation (5) has a root between .7 and .8 ; hence equation 
(6) has a root between and .1. 

By trial of hundredths, trying .02, .03, .04 ••• it is found that 
this equation has a root .03 + , between .03 and .04 and 
evidently nearer .03. 

Hence our original equation has a root 4.73 + . 

In this way we can compute any real numerical root of a 
rational integral algebraical equation to any desired number 
of significant figures. 

Illustrative Problems. — 1. Find the cube root of 
1,624,276 to four significant figures. 



x 3 — 1,624,276 = 0. By trial, substituting 100, 200, 
found to have a root between 100 and 200. 



for x, this is 



x - 100) 
100) 



x' - 10) 
10) 



7) 
7) 



1 + + - 1,624,276 

■+ 100 + 10000 - 1,000,000 
1 + 100 + 10000 (- 624,276 

+ 100 +20000 
1 + 200 ( + 30000 

+ 100 

1 + 300 + 30000 - 624,276 

+ 10 + 3100 - 331,000 
1 + 310 + 33100 (- 293,276 

+ 10 + 3200 
1 + 320 (+36300 

+ 10 

1+330 + 36300 - 293,276 

+ 7 + 2359 + 270,613 



1 + 337 

+ 7 



+ 38659 (- 
+ 2408 



22,663 



1+344 (+41067 

+ 7 

1 + 351 + 41067 - 22,663 



By derivation, the roots 
are 100 less than the roots of 
(1) ; hence a root between 
and 100. By trial, substitut- 



By derivation, has a root 
between and 10. By trial, 
a root between 7 and 8. 



By derivation, root between 
and 1. By trial, between 
.5 and .6, and nearer to .5. 



Hence the root of the original is 117.5, which may be partially checked 
by four-place logarithms. 



NUMERICAL ALGEBRAIC EQUATIONS 403 

2. Compute one negative root of 

2x 4 + 10a 3 - Sx 2 - 11a + 19 = 0. 

Negative root between — 1 and — 2. 
x + 2) 2 + 10 - 8 -11 +19 

-2) - 4 - 12 +40 - 58 

2+6-20 +29 (-39 
_ 4 - 4 +48 
(+77 



2+2 
- 4 


-24 

+ 4 


2-2 
- 4 


(-20 



x _ .6) 2 — 6 - 20 +77 - 39 Root between and 

+ .6) + 1.2 - 2.88 -13.728 +37.9632 1. By trial, be- 
2 + 4.8 - 22.88 + 63.272 - (1.0368 tween .6 and .7. 

+ 1.2 - 2.16 -15.024 
2_ 3.6 -25.04 (+48.248 

+ 1.2 - 1.44 
2- 2.4 (-26.48 

+ 1.2 
2 - 1.2 - 26.48 + 48.248 - 1.0368 By derivation has a 

root betwen and 
.1. By trial, be- 
tween .02 and .03. 
The original equation has a root — 2 + .62+, or — 1.38 - , i.e. between 
- 1.38 and - 1.37. 

PROBLEMS 

See the two preceding sets of problems and use the results 
obtained. 

1. Compute to three significant figures the largest positive 
root of the following equations, 

a. 2 a? + 3 x 2 - 9x - 7 = 0. 

b. 2 X s + 3 x 2 - 7 = 0. 

c. 2^-9^-7 = 0. 

d. x 4 - 2 x 3 + 3 x 2 - 18 x + 21 = 0. 

2. Compute by the process indicated the positive root of 
x? — 3 x — 21 = to three decimal places ; compute the same 
by solving as a quadratic, and compare as to efficiency the two 
methods. 



404 UNIFIED MATHEMATICS 

3. Solve the equation 4 a 3 — 3x + .5= 0, computing the 
smallest positive root to four decimal places. This is the 
value of sin 10° ; check by your table of sines. 

4. In problem 5, page 245, you have computed the sine of 
3° to four decimal places. Write the cubic which will give the 
sine of 1° ; compute the smallest positive root, and discuss to 
what decimal place it could be carried with propriety when 
the sine of three degrees is given to four decimal places. 

5. Plot the graphs of the two equations { ' 

Note that the points of intersection give the solutions of the 
two equations regarded as simultaneous ; but solving the two 
equations as simultaneous equations, we are led by substitution 
to the cubic x • \ (x 2 — 16) = 1, or x 3 — 16 x — 4 = 0. Solve the 
cubic and compare with the solutions obtained graphically. 

6. Historical problem. The great Archimedes proposed the 
problem to cut a sphere by a plane in such a way that the 
two segments of the sphere should have to each other a given 
ratio. Archimedes showed that the solution could be obtained 
as the intersection of a hyperbola and a parabola. If the 
diameter of the sphere is taken as 10 and k as the ratio of the 
larger to the smaller segment, this problem leads to the cubic 

x * _ 300 x + 2000 ft- 1 ) = 0, 
fc-h 1 

in which x represents the distance of the plane from the center 
of the sphere. Solve to two decimal places when k = 2. The 
plane at a distance x from the center then trisects the sphere. 

7. In the preceding problem show that the solution may be 
obtained as the intersection of a hyperbola and a parabola. 

8. A famous problem of antiquity is the problem to dupli- 
cate a given cube, i.e. to solve geometrically x 2 = 2 a 3 , a being 
the side of the given cube. Long before analytical geometry 
was invented it was known that the solution could be given as 
the intersection of the parabola x 2 = ay with the parabola 



NUMERICAL ALGEBRAIC EQUATIONS 405 

?/ 2 = 2 ax. Construct the graphical solution when a is taken 
as 10. 

The problem may also be solved, by the intersection of either 
of the two given parabolas with the hyperbola xy=2 a 2 . Verify. 

If two means, x and y, are inserted between a, and 2 a, i.e. 

- = - = i, then x is the solution of the equation x z = 2 a 3 . 
x y 2 a 

This method reduced the problem of the duplication of the 

cube to the problem of inserting between two given numbers, 

or lines, two geometric means. 

9. The volume of a spherical segment, greater than a hemi- 
sphere, of height x + r, is given by the expression 

F = ?(2)H3r 2 x-x 3 ); 
o 

the volume of a sphere is f ^r 3 . Find the segment of a sphere 
of water of radius 10 which will be equal in weight to a 
sphere of wood, radius 10, which wood is only .6 as heavy as 
water. This leads to the cubic equation 

.6(| Trr 3 ) = - (2 r 3 + 3 r 2 x - x*), 

or x 3 — 3 r 2 x + .4 r 3 = 0, 

and r + x is the depth to which the sphere of wood will sink 

when it is placed in water. Compute this depth when r = 10. 

10. Ice is only .92 as heavy as water. Use the equations of 
the preceding problem, substituting .92 for .6, to find the depth 
to which a spherical iceberg of radius 100 feet, if one were 
possible, would sink in water. 



406 



UNIFIED MATHEMATICS 




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CHAPTER XXVI 

WAVE MOTION 

1. General. — In nature there are two types of recurrent 
motion, somewhat closely connected mathematically, in which 
repetition of motion occurs at regular intervals. 

One type of this motion, in cycles as we may say, repeats 
the motion in one place, and is in a sense stationary. The 
tuning fork in motion moves through the same space again 
and again ; a similar movement is the motion of a vibrating 
string. Of this stationary type may be mentioned the heart- 
beats, the pulse, the respiration, the tides, and the rotation 
of a wheel about its axis. 

The second type of recurrent motion transmits or carries 
the vibratory impulse over an extent of space as well as time. 
The waves of the sea are of this character. Sound waves, 
electrical vibrations or waves, and radiant energy vibrations 
are transmitted by a process similar to that by which the 
waves of the sea are carried. 

Both of these types of motion are representable mathe- 
matically by equations involving a sequence of trigonometric 
functions. To the fundamental and basic function involved, 
y = sin x, we will direct our attention in the next section and 
to simple applications in other sections of this chapter. 

2. The sine curve. — As a radius vector of unit length ro- 
tates in a plane with uniform velocity about a center, the sine 
of the angle 6 fluctuates between 1 and — 1. The variation of 
sine 6 may be represented by the movement on the ?/-axis 
of the projection of the vector, and this movement of the 

407 



408 



UNIFIED MATHEMATICS 





, ^ „ 


-.*■* ^ ,*^"" ^s 


-7 s S v 


J \ / V 


-T V / \ 


4 X-A \ 


1 k_ _^>_ u 




\ \A -iM S v ^' 


\ v s y 


S n t n„ ^/ 


^ ^Z S s» ^ 




'»■_ — -' »= — 





Graph of i/ = sin ; a pure sinusoid 

The length AA' equals the circumference of the circle ; the amplitude, 
vertical distance between highest and lowest points, equals the diameter 
of the circle. 




Graph of y = sin 2 0; a sinusoidal curve 

The frequency is double that represented in the preceding graph. 




y = sin0 + sin 2 0; obtained by addition of corresponding ordinates in 
the two preceding curves 

This type of curve is obtained from a tuning fork having an octave 

overtone. 



WAVE MOTION 409 

projection is termed simple harmonic motion, frequently 
abbreviated S. H. M. Precisely the same type of movement 
is given by the projection of the moving vector on the ce-axis, 
x — cos 0, or on any line in the plane of the motion, 

Z = COS (0 — e), 

wherein e is the slope angle of the line and z is the projection 
of the radius. 

If the vector completes one revolution, 2 7r r , or 360°, in 1 
second, the period of the motion is called 1 second, and the 
frequency, or the number of repetitions of the complete move- 
ment or cycle in a second, is 1 per second. If the complete 
revolution is effected in i second, the period is i second and 
the frequency 2 per second. The graphs of y = sin and of 
y = sin 2 represent under these conditions the progress of the 
ordinate for uniform changes in 0, i.e. for uniform changes 
in the time, since the rotation is with constant angular ve- 
locity. For convenience the angle is conceived as measured 
in radians and the radius is taken on the x-axis as the unit to 
represent one radian ; the abscissa then corresponds either to 
the angle measured in radians or to the length of arc traversed 
by the end of the moving vector. In plotting y = sin 0, values 
of 6 from to 360° or from to 2 7r r are plotted on the horizon- 
tal or 0-axis. Note particularly the points for which 0=0, 
30°, 45°, 60°, 90°, ... 180°, ... 360° ; or 

^ if if 7T r 7T r o . 

Note that AA' on our diagram represents one complete cycle 
or period. For many purposes it is desirable to take t, the 
time (in seconds, usually), as the variable. The same graph 
then represents y = sin 2 wt, wherein AA' is taken equal to 1 and 
the horizontal axis is the Z-axis. The same curve represents 
y = sin 20 -n-l, if A A' is taken as ^ of 1 unit of time. The 
upper curve in our diagram is a pure sinusoid, the distance AA' 
representing the circumference of the circle of which the 
maximum ordinate is the radius. 



410 



UNIFIED MATHEMATICS 



The two curves plotted should be carefully studied; the 
lower curve has double the frequency of the upper and one 
half the period. The swing, amplitude as it is termed, is the 
same ; the amplitude is the algebraic difference between the 

maximum and minimum values of 
the function. 

Any curve representing 

y = a sin or a sin £0 
or a sin (£0 -f e) 

is called a sinusoid. We shall find 
that the graphs of y = a cos 9, 
y = a cos kO, and y = a cos (kd + c) 
differ from the preceding only in 
position. 

For most purposes it is conven- 
ient to plot time in complete units 
on the ordinary coordinate paper, 
the unit depending on the period 
of time in question. For a com- 
plete rotation in one minute ten 
seconds might be taken as one 
unit on the horizontal axis with 
the radius as vertical unit, and the 
curve would differ very slightly 
from our curve. The highest and 
lowest points would fall then at 
15 and 45 respectively ; 0, 5, 7.5, 
10, 15, 20, and 30 seconds corre- 
spond then to 0, 30°, 45°, 60°, 90°, 
120°, and 180° respectively. 

Physicists and engineers com- 
monly draw the sinusoidal curves 
which are of frequent occurrence entirely from graphical 
considerations. The circle with the desired amplitude is 




S 



WAVE MOTION 411 

drawn ; the angle between the axes is bisected and re- 
bisected (as often as desired) ; an appropriate length for a 
complete c} T cle is taken on the horizontal axis, and this is 
divided into jnst as many parts, usually 16, as the circum- 
ference of the circle is divided by the axes and the bisecting 
lines which were drawn. At each point on the horizontal axis 
an ordinate is drawn and from each corresponding point on 
the circle a horizontal line is drawn to intersect the correspond- 
ing ordinate. Corresponding points have the same numbers 
if on the circle intersection points are numbered from the 
right-hand intersection with the horizontal axis counter-clock- 
wise and numbered on the line from the left-hand end of the 
horizontal length taken to represent the time of one cycle, as 
indicated on our diagram. The two upper figures, page 408, 
were drawn by this method. The student is urged to make both 
the graphical construction and the construction by using the- 
numerical values of the sines from the tables. Compare also 
the work under Section 11, Chapter VII. 

PROBLEMS 

1. Plot the curves y = shi$° and y — sin 3 0° on the same 
sheet of coordinate paper ; take 1 inch as radius and on the 
horizontal axis take 1 inch to represent 60°. For a pure 
sinusoid, y = sin x, one unit on x should be the length of the 
radius ; then 3.14+ radians represents 180°, the second point 
in which the curve y = sin x cuts the axis of abscissas. 

2. Plot y = sin 2 irt\ note that t = ^, -fo, ••• corresponds to 
36° and multiples ; take one unit for i as 6 times the radius 
chosen. 

3. How could you interpret the curve of the preceding exer- 
cise as y = sin 4 -n-t ? 

4. Plot 10 points of y = sin (0 — 30°). This curve is similar 
to the preceding; it is 30° behind, we may say, the regular 
sine curve ; the " lag " is 30° ; the two curves y = sin and 



412 UNIFIED MATHEMATICS 

y — sin (0 — 30°) are said to be out of phase, the phase angle 
of the second being — 30°. The " phase angle " is of particu- 
lar importance in the theory and practice of alternating cur- 
rents. 

5. Plot the curve E = 110 sin 0. Note that if the horizon- 
tal scale be taken so that 1 inch represents 60° and the vertical 
scale such that 1 inch represents 110, the curve is precisely 
the first curve of problem 1. This curve represents the vari- 
able electromotive force (e. m. f.) developed by a generator 
which generates a maximum e. m. f . of 110 volts. To plot 
the curve no knowledge of electricity is necessary, but com- 
plete interpretation requires technical knowledge. 

3. Sound waves. — If a tuning fork for note lower C is set 
to vibrating, the free bar makes 129 complete, back-and-forth, 
vibrations in one second. By attaching a fine point to the end 
of the bar and moving under this bar at a uniform rate, as it 
vibrates, a smoke-blackened paper, a sinusoidal curve is traced 
on the paper. Our curve is traced by a bar vibrating 50 times 
in 1 second. 




The curve y = sin (50 X 2 irr) 

Tuning fork vibrations recorded on smoked paper. 

In 1 second 50 complete vibrations are made ; the vertical 
distance between the top and the bottom of the arcs repre- 
sents the distance moved by any point on the moving bar ; 
the motion is simple harmonic (S. H. M.). The period is -^ 
second ; the "frequency is 50 ; the amplitude is about -g 1 ^ inch. 
If the smoked paper were moved with uniform velocity under 
the vibrating bar in such a way as to cover 50 times the cir- 
cumference of a circle with radius -^ of an inch, or 50 x 2 tt X -%\ 
inch per second, the curve traced would be almost a perfect 
sinusoid of the type y = sin 0. The points move of course on 



WAVE MOTION 



413 



arcs of curves, but the variation from a straight line is ex- 
tremely slight. 

Corresponding to each movement of the vibrating rod there 
is a movement of the air. As the bar moves to the right it 
compresses the layer of air to its right and that compression 
is immediately communicated to the layer of air to the right ; 
as the bar moves back and to the left, the pressure on the ad- 



ou 




Vibration records produced by the voice 
"a" as in "ate" ; "ou " as in "about " ; "r ,? in "relay " ; "e" 
in i; be" ; and "a" in "father.'" The tuning fork record, frequency 
50 per second, gives the vibration frequencies. 



jacent air is released and a rarefaction takes place. In -^ of 1 
second you have the air adjacent to the rod compressed, back 
to normal, and rarefied; during this time the neighboring air 
is affected and the compression is communicated a distance 
which is the wave length of this given sound wave. In 1 second 
this disturbance is transmitted 1100 feet at 44° Fahrenheit. 
The wave length for this sound wave then is ~\^- = 22 feet. 

The wave length is commonly designated by A. If v is the 
velocity, and t the time of one vibration, A. = vt. 



414 UNIFIED MATHEMATICS 

The notes of the key of C on the natural scale have the fol- 
lowing vibration frequencies : 

cde f g a be' 

256 288 320 341.3 384 426.7 480 512 

The intensity or loudness corresponds in the rod to the 
length of swing of the vibrating rod; as this amplitude de- 
creases, the intensity of the sound decreases. For small am- 
plitudes the vibratory motion gives a convenient way of 
measuring small intervals of time. 

Thus on the above diagram if the tone of the note lower C, 
y = sin 256 irt, were represented, each complete wave would 
represent T \-% of 1 second ; each half or each arch would rep- 
resent 2-lg- of 1 second. Tuning bars, with periods -^ and 
yi^ of 1 second are run electrically for timing purposes. 

The curve y = sin 6 -f sin 2 6 represents the combination in 
sound of two tones which differ by an octave. Precisely the 
type of curve which is represented by our diagram can be 
produced mechanically by the record of a vibrating tuning 
fork 1 which sounds not only the principal note but also the 
octave overtone, due to the fact that the bar vibrates about the 
middle point at the same time that it vibrates about the end. 
Vibrating strings also have multiple vibration, overtones and 
other tones. Harmony is the result, in general, when the 
vibrating instrument gives vibrations which are connected 
with the fundamental vibration by simple numerical relations, 
like that of the overtone. 

Thus the notes of the major chord, key of C, c, e, g, c, on 
the piano, have the vibration frequencies in the ratios 4 to 5 
to 6 to 8. 

4. Helical spring. — Similar to the vibrations of the air are 
those of a spiral wire spring which oscillates back and forth 
when a weight is suspended by the spring ; the successive 
compressions and elongations of the wire correspond quite 

1 See Miller, The Science of Musical Sounds, p. 188, for photograph. 



WAVE MOTION 415 

closely to the condensations and rarefactions of the air. The 
position of the weight at any instant can be given by an 
equation entirely similar to the equation above of note C. 
Thus if the time of one complete vibration is \ second, and 
the maximum displacement is 4 inches, the equation is 

?/ = 4 sin 4 irt r . 

This gives the elevation above and below the point at which 
the weight comes to rest. Perfect elasticity of the spring is 
assumed. 

5. Light waves. — Light waves have a much higher velocity 
than sound waves, 300 x 10 6 meters per second. The different 
wave lengths correspond to different colors, just as different 
wave lengths in sound waves correspond to different tones. 
The wave length of the light from burning sodium (D 2 of the 
spectrum) is 0.5890 x 10~ 6 meters per second, and for other 
colors varies for the visible spectrum between .39 and 
.75 x 10" 6 meters. The vibration frequency of the sodium 
light is the number of these waves which occur in one second 
of time, hence since these waves cover 300 X 10 6 meters in one 
second the frequency n is such that 

n • A = v, or n • 0.589 x 10" 6 = 300 x 10 6 , 

whence n = = 509 x 10 12 vibrations per second. 

.589 x 10- 6 r 

Radiant energy is of the same general nature with longer 

waves. Light waves differ from the sound waves in having 

transverse vibrations, not longitudinal. 

6. Electricity. — In electricity, particularly in the discussion 
of alternating currents, the sine curve plays a prominent role. 

The equations e = 156 sin 0, 

i = 4 sin 0, 
and p = el = G24 sin 2 6 = 624(1 _ i cos 2 6) 

= 312 -312 cos 2 0, 



416 



UNIFIED MATHEMATICS 



represent respectively the electromotive force, e, measured in 
volts, and the current, i, measured in amperes, and the power, 

p, measured in watts 
of an ordinary electric 
current. 

In general, current 
and electromotive 
force are " out of 
phase " ; the equa- 
tions when the cur- 
rent lags 30° behind 
the electromotive 
force are. 




e = 156 sin 0, 

« = 4 sin(<9-30°). 



Sinusoids traced by electrical means 

Oscillogram of an alternating current in which 
current and e. m. f. are " in phase.' ' 

On the two diagrams 
we have represented by a photographic process the magnitude 
of the current and electromotive force of an alternating cur- 
rent. The current is represented by the curve with the 
smaller amplitude. 
In the first illus- 
tration current and 
e. m. f. are " in 
phase," and under 
these conditions a 
maximum of power 
is developed ; in the 
second illustration 
current and e. m. f. 
are " out of phase," 
the current lagging 
behind the e. m. f . 

The power at any instant delivered by an alternating current 
is given by the product of the current and the e .m. f. at that 
instant. Employing the formulas, 




Oscillogram showing current curve (lower) lag- 
ging 90° behind e. m. f. curve 



WAVE MOTION 417 

cos (« — /?) = cos a cos (3 + sin a sin /?, 
cos (« 4- (?) — cos a cos /? — sin a sin /?, whence 
cos (a — /?) — cos (a + /?) = 2 sin a sin /?, 
show that p=ie sin sin (0 — 30°) may be reduced to 

*i[cos 30° - cos 2(0 - 15°)]. 

A 

Plot the curve showing the power at any instant, when 

e = 156 cos 6 and i = 4 cos (0 - 30°). 

Note that this power curve is also a sinusoidal curve but placed 
with reference to a horizontal line which runs 270 units above 
the rc-axis. 

PROBLEMS 

1. Plot the curves y = sin 256 it t r and y = sin 512 ?r t% 
using \ inch for 1 on the vertical axis and 6 half-inches for yjg- 
of 1 second on the t or horizontal axis. Treat the equations 
as y = sin 2 7r£ r , and y = sin 4 7rt r , respectively, substituting for 
t, 0, .1, .2, .3, — , .9, and 1 instead of ^ of T -Lg, -^ of T -|^, •••. 

Note that the unit T |-§ taken as 1 on the horizontal axis, disposes of the 
difficulty of the awkward fractions. 

2. What is the frequency of the vibrations in the curves 
of the preceding example ? What are the corresponding wave 
lengths ? 

3. How would y = cos 2 nf differ from the curve for 
y = sin 2 irt r ? AVrite 10 values of y = cos 2 >rrt r for t = 0, 1 L ) 

1 1 1 1 5 1 ... 1 

8"> 6"? 4' ~3> T"2> 2"' X " 

Note that these angles correspond to 0°, 30°, 45°, 60°, ••• respectively. 

4. Use the equation cos = sin (90° + 0) to show that 
y — sin lags 90° behind y = cos 0. 

5. Draw the graphs of y = cos and y = cos 2 0; divide the 
arc of the circle into 24 equal parts and take the distance rep- 
resenting 2 if as divisible by 24. 



418 UNIFIED MATHEMATICS 

6. Draw the graph of y — sin ( - + $ ) and compare with 

y = sin 6. Discuss the corresponding motion of the moving 
point on the vertical axis. 

7. The limits of hearing are for vibrations of 16 per second 
and 40,000 per second. What are the corresponding wave 
lengths ? 

8. Plot on the same diagram the two curves, 

e = 156 sin 6, 

i = 4 sin (0 - 30°). 

9. In problem 8 find the value of e for each 30° to 360°. 
This completes a " cycle " of values. The time of this move- 
ment in a 60-cycle system is -^ of 1 second. What is the 
value of t for the angles given, and also for = 45°, 135°, 225°, 
and 315° ? 

10. On the curve, on the same axes as the preceding, 
i = 4 sin (0 - 30°), read the values of for theiangles 30°, 45°, 
60°, 90°, --to 360°. These may represent current in the cir- 
cuit of problems 5 and 6 ; the current lags 30° behind the 
e. m. f. What interval of time is represented by the 30° lag ? 

11. Plot to the same axes the curves, i = 4 sin (6 + 40°), 

e = 156 sin $. 

The curve of i here leads the curve of e by 40°. 
In the case of i, what are convenient values of to plot 
without using tables ? 

12. Assuming that it takes ^ of 1 second for one com- 
plete cycle of i or e in problem 8, find the time difference 
represented by the 40° angular difference. Find angles ap- 
proximately corresponding to T ^, Y \^, j^, T ^, and ^ of 
1 second. 

7. Sine curve ; circle ; ellipse ; cylinder. — If a circular cyl- 
inder, such as the one in our diagram, is cut by any plane, the 



WAVE MOTION 



419 



intersection is an ellipse. 
Thus, the plane through 
AOB in our diagram, in- 
clined at an angle of 45°^ 
cuts the cylinder in an 
ellipse whose equation is 

a 2 2 a 2 

The circular base and any 
parallel section has the 
equation 



+ ^ = 1 - 

a 2 



Taking the portion of the 
cylindrical surface be- 
tween the elliptical curve 
and the circular curve 
through the same center 
and unrolling it gives a 
pure sinusoid, 

y = a sin 0. 

In the figure PQ = QM t 
since the cylinder is cut 
at an angle of 45°. But 
QM, the ordinate on the 
circle, equals a sin and 
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v — "^-J^. 


M 




■^-^ 


■ill 


; 


-/ :- 






\ 


- V 




















f 


















-T Lf i I 




■ ■ 














■ 














i 
























































■ii ' • , 
















































__ 






ill. 


1 




.""" 


r— 


lllj 


— ^a. 






























s 




























































^L 










J 




V 










1 1 / ■ . | ! 




V 










1*1 1 1 




v. 










^S 












































, 


Ill, 














A plane intersecting all the elements of 

a circular cylinder cuts the surface 

in an ellipse 

the arc AQ, which will be the ab- 
scissa, is ad. 

Hence the curve, when 
rolled out, is 



Sinusoid developed by means of 
cylinder 



y = a sin v 

and will give a complete 

arch of the sinusoid if 

circular U PP er and lower Portions 

of the surface are given. 



420 



UNIFIED MATHEMATICS 



Experimentally the student can develop this curve by roll- 
ing a sheet of paper about a cylinder and cutting out with a 
sharp knife the required portions. 



i- 



I 



ffi¥&E±=8L 



Piston-rod diagram 
AB, the stroke ; H C, connecting rod ; OC, crank arm. 

8. Piston-rod motion. — The common piston-rod motion of 
engines furnishes abundant trigonometric material, much of 
which is of sufficiently elementary character so that by the 
application of simple formulas problems of interest to the 
engineer can be solved. 

The essential features for our purposes are the piston head 
H, the connecting rod HO of length I, the crank arm 00 of 
length r, and the stroke AB, which is the distance through 
which the piston head // moves. Were the connecting rod 
infinite in extent, the motion of H would be simple harmonic 
motion when C is rotating with uniform velocity about 0. 

In modern engines the ratio of I to r varies from 3 to 1, low, 
to 4.8 to 1, which is approximately that of a Ford engine. It 
is desired to find for each position of the piston head the 
angle a of the connecting rod, the angle of the crank shaft, 
and also the effective pressure, called the tangential compo- 
nent, of the connecting rod to turn the crank shaft. 

In the first place, when I : r = 4.8 : 1, the angle a never 

exceeds arc sin — -• Determine this angle in degrees. As the 
4.8 

pressure P at H is horizontal, only a portion, Pcos a, of this 



WAVE MOTION 



421 



pressure is communicated to the connecting rod. Discuss the 
variation in pressure due to the inclination of the connecting 
rod and note that it is relatively small. Of course the pressure 
of the gas in the cylinder chamber is not uniform and this 




Connecting rods and crank arms in six-cylinder automobile engine : 
ratio l:r = 3:1 



variation is much more serious than the variation due to the 
angle of a connecting rod. 

Find also the maximum vertical pressure P sin a on the 
cross-head support. 

Show that the length of the stroke AB is equal to the 
diameter of the crank circle. 

The position of the piston head is indicated in decimal parts 
of the total stroke 2 r, as measured from A. In our diagram 
the piston head is at .75 of the stroke. To determine a and 
when the position of the piston head is given we solve a 
triangle in which the three sides are known. Thus on our 
diagram 00'= 4.8 ; HO = 4.3 ; HC= 4.8 ; and 00= 1. Solve 
this triangle and determine the angles a and 6. Commonly a 
diagram is drawn in which the angles 6 of the crank arm are 
plotted as abscissas and the piston displacements are plotted 
as ordinates. 



422 



UNIFIED MATHEMATICS 



PROBLEMS 

1. Given (9 = 0, 10°, 20°, 30°, 60°, 90°, 120°, 150°, and 180°, 
find the corresponding values of a and the positions of H as 
decimal parts of the stroke. Plot angles as abscissas and 
piston displacements as ordinates. How could you complete 
this to 360°? Could you go on beyond 360°'/ Take l:r = 4.8:1. 

2. Assuming that the piston rod transmits a uniform pres- 
sure of 200 lb., find the effective turning pressure when 
6 = 30°, 60°, and 90°. Resolve the force at C into two com- 
ponents, one normal to the crank shaft and the other along 
the crank shaft. The component normal to the crank shaft, the 
tangential component, is effective, i.e. does the work. The 
radial pressure is also computed and is used to determine 
friction loss. Find the values of the radial pressure corre- 
sponding to above tangential pressure. 

Compute in this case for the given angles the pressure on 
the cross-head support by the connecting rod due to the com- 
ponent, P sin a. 




Connecting rod, crank arm, and cylinder on A. T. & S. F. locomotive 1444 

The stroke is 30 inches and connecting rod 60 inches. 



3. Draw the graph illustrating relative positions of the 
piston head and the connecting rod when the crank-pin is at 
the lowest point, in the locomotive illustrated above. 

4. Find the number of strokes of the piston per minute 
when the train moves 60 miles per hour, given that the driving 
wheels are 57 inches in diameter. 



CHAPTER XXVII 

LAWS OF GROWTH 

1. Compound interest function. — The function S = P(l+ i) n 
is of fundamental importance in other fields than in finance. 
Thus the growth of timber of a large forest tract may be ex- 
pressed as a function of this kind, the assumption being that 
in a large tract the rate of growth may be taken as uniform 
from year to year. In the case of bacteria growing under 
ideal conditions in a culture, i.e. with unlimited food supplied, 
the increase in the number of bacteria per second is propor- 
tional to the number of bacteria present at the beginning of 
that second. Any function in which the rate of change or 
rate of growth at any instant t is directly proportional to the 
value of the function at the instant t obeys what has been 
termed the " law of organic growth," and may be expressed by 
the equation, 

y = ce kt , 

wherein c and k are constants determined by the physical facts 
involved, and e is a constant of nature analogous to ir. The 
constant k is the proportionality constant and is negative when 
the quantity in question decreases ; c is commonly positive ; 

e = 2.178 .». 

The values of the function of x, ce kx , increase according to 
the terms of a geometrical }rrogression as the variable x increases 
in arithmetical progression. 

2. ir and e. — A function can be found by methods of the 
calculus which is such that the rate of growth of the function 

423 



424 



UNIFIED MATHEMATICS 



at any instant t, or x, exactly equals the value of the function at 
that instant. This function is given by the equation, 

y = e l or y = e x . 

The constant e is represented by the series 

e = l + t + u: + i_s + l£** , > 

wherein [3, called factorial 3, represents 3x2x1, and 
[4 = 4x3x2x1, and, in general, \n, n being a positive in- 
teger, represents the total product of all the integers from n 
down to 1. The sum of this series to 5 decimal places is 
2.71828 ; to 10 places, = 2.7182818285. 

The curve represented by the 
equation 

y = e x 

is such that at any point on this 
curve the slope equals in nu- 
merical value the ordinate at 
that point. 

The graph of y = e~ x is such 
that the slope at any point is 
the negative of the ordinate at 
that point. 

The graph of y = e kx is such 
that the slope at any point is k 




Graphs of y = e x and y = 10* 

times the corresponding value of the function at that point. 
Values of the function y = e x may be determined by loga- 
rithms. 



LAWS OF GROWTH 425 

Thus to find the points on y = e x , for which 

x = — 2, — 1, 0, .1, .2, .5, .8, 1 } 2, and 3 respectively, 
we take first log y = x log e ; since log e = .1343 



X — - 


-2, 


log,?/ = 


- .8686, 
9.1314 - 


-10 


y = 


.135 


X = - 


-1, 


log y = 


- .4343, 
9.5657 - 


-10 


y = 


.368 


x = 


o, 


log ?/ = 


o, 




y = 


1 


X = 


.1, 


log 2/ = 


.0424, 




y = 


1.103 


a = 


2 


log 2/ = 


.0869, 




y = 


1.222 


# = 


.5, 


log ?/ = 


.2171, 




y = 


1.649 


a; = 


.8, 


log 2/ = 


.3474, 




y = 


2.225 


x = 


1, 


log?/ = 


.4343, 




y = 


2.718 


a? = 


9 


log y = 


.8686, 




y = 


7.390 


x = 


3, 


log y = 


1.3029, 




y = 


20.09 



Similarly, if y = ce kx , log y — log c + kx log e, and these values 
are obtained by logarithms. 
The limit of the expression 



1 + 



U 



as n approaches infinity gives the value, e. When n is taken 
as a large positive integer, it can readily be shown that this 
expression . _, N n 

has a value differing but slightly from e. 

e and ir may be called fundamental constants of nature ; in 
mathematical work as applied to statistics and to physical 
problems of varied kinds these constants often appear. 

3. Natural logarithms. — The first logarithms as computed 
by Napier were not calculated to any base, but were founded 



426 



UNIFIED MATHEMATICS 



upon the comparison between an arithmetical and a correspond- 
ing geometrical progression. However, a base of the Napierian 

logarithms can be established, and it is approximately -. 




Graph of y = log e x 

The ordinates represent the "natural logarithms 
of the numbers represented by the abscissas. 



From the mathematical point of view, and 
indeed for many applications of mathe- 
matics to physical problems, the base e is 
preferable to 10 as the base of a system of 
logarithms. Logarithms to the base e are 
called natural logarithms. 



4. Application. — The most immediate application of a 

function in which the growth is proportional to the function 

itself is to the air. The decrease in the pressure of the air 

at the distance h above the earth's surface is proportional 

to li. 

h 
The expression P=760 e 7990 gives the numerical value of 

the pressure in millimeters of mercury for h measured in 

meters. The negative exponent indicates that the pressure 



LAWS OF GROWTH 427 

decreases as h increases. In inches as units of length of the 

mercury column, h in feet, 

h 
P = 29.92 e~262oo. 

This is known as Halley's Law. 

The growth of bean plants within limited intervals and the 
growth of children, again between quite restricted limits, 
follow approximately the law of organic growth. Radium in 
decomposing follows the same law, the rate of decrease at any 
instant being proportional to the quantity. In the case of 
vibrating bodies, like a pendulum, the rate of decrease of the 
amplitude follows this law ; similarly in the case of a noise 
dying down and in certain electrical phenomena, the rate of 
decrease is proportional at any instant to the value of the 
function at the instant. 

PROBLEMS 

1. By experiment it has been found that 1000 of the so- 
called "hay bacteria" double their number, under favorable 
conditions, in 20 minutes. Find the rate of growth per 
minute. Take n __ 1000 e kt 

and determine k by substituting n = 2000, t = 20, and solving 
for k. Determine the number that would grow from 1000 
bacteria in 1 hour ; in 1 day. 

2. The cholera bacteria have been found, under favorable 
conditions, to double their number in 30 minutes. Determine 
the rate of growth per minute, and the number that would 
grow in one day from 1000. 

Note . The favorable conditions cannot be continued for such a period. 

3. Assuming that 

P=760e~^ 

find the value of h which will reduce the pressure of the air by 
1 mm. Take the logarithm of both sides and note that 



428 UNIFIED MATHEMATICS 

— — — log 10 e must reduce log 760 to log 759. Find at what 
8000 

height the mercury column is reduced 1 cm. At what height 

would the pressure be reduced to 660 ? Are these heights 

ever attained ? 

4. Find the barometric pressure when 

h = 1000, 5000, 10,000, and 15,000 feet, assuming 

P= 29.92, when h = 0, in which P is the numerical value 

of the pressure in inches of mercury and the height is h feet. 

Find the height for which the pressure decreases ^ of 1 inch. 

5. Show that if the height of the elevation is measured in 

miles, the pressure in inches is given approximately by the 

formula h 

P = 29.92 e~5. 

Note that 26,000 is nearly 5 x 5280. The constant 26,200, above, is 
taken for simplicity instead of 26,240. 

6. To what change in height does the maximum variation 
of the barometer recorded on the photograph, page 60, 
correspond ? 

7. Compute by the progressive method of Section 4, 
Chapter XII, the value of e, from the series, 

= l+l + L i + i + L i + ,i + L i + i + L i+"- 
by summing 8 terms to 8 decimal places. 

8. Plot on the same diagram and compare the two graphs, of 

y = e x and y = 10*. 

9. Plot the curve y = e~ x \ taking J inch as .1 on the hori- 
zontal and on the vertical axes. This curve represents what 
is termed the normal distribution curve, which is of funda- 
mental importance in all statistical work. In general, large 
groups of individuals may be distributed as to ability in any 
given quality over the area under such a curve ; the middle ab- 
scissa at x = represents average ability, and deviation to one 
side or the other represents, on one side, ability above the aver- 



LAWS OF GROWTH 



429 



age and, on the other, ability below the average. The total 
number of all individuals considered is represented by the 
area between the curve and the ce-axis. 

An interesting graphical illustration of a normal distribu- 
tion curve is the crowd at a football game when a great 
bleacher is not filled. The central aisles are all filled to a 
height representing the middle ordinate and from this out in 
either direction, the ordinates drop off, frequently in strikingly 
symmetrical manner, and corresponding quite closely to the 
normal distribution curve. 

5. The curve of healing of a wound. — Closely allied to the 
formulas expressing the law of organic growth, y = e kt , and the 
law of " organic decay," y — e~ kt , is a recently discovered 
law which connects algebraically by an equation and graph- 
ically by a curve, the surface-area of a wound, with time 




DayO 4 8 
May lb 19 23 



56 60 64 66 
,10 14 18 20 



Progress of healing of a surface wound of the right leg, patient's age 31 years 

The observed curve oscillates about the smoother, calculated curve. 



expressed in days, measured from the time when the wound is 
aseptic or sterile. When this aseptic condition is reached, 
by washing and flushing continually with antiseptic solutions, 
two observations at an interval commonly of four days give 
the "index of the individual," and this index, and the two 
measurements of area of the wound-surface, enable the physi- 
cian-scientist to determine the normal progress of the wound- 
surface, the expected decrease in area, for this wound-surface 
of this individual. The area of the wound is traced carefully 



430 



UNIFIED MATHEMATICS 



on transparent paper, and then computed by using a mathe- 
matical machine, called a planimeter, which measures areas. 

The areas of the wound are plotted as ordinates with the 
respective times of observation measured in days as abscissas. 
After each observation and computation of area the point so 



Sq.cm. 

Area 

10 



V 












1 


4 


















I 


1 1 






S 


o 








* 




t 
















i 


,,/irir 


na 


I Index = 


= '). 


14 




\ 


\ 


-" 










> 




- 


,U 


""■°v 
























y 








V-c 
\ 


,<4y 






t 


T 








\- 




















\ 








\ 






\L 






































s 


s 








.?■ 




























































t 




































^ 


N 












> 




































Y** 












r 




























'•; 
















































X 




















TT-- ? >»> ■ 



















Observed 8 
Calculated 

Sept. 23 Od. 1 



Dai 



1G 24 32 40 
8 16 24 32 40 44 

8 12 



9 17 25 Nov. 2 

Progress of a surface wound of the right knee 
Two infections in the course of healing are indicated 



obtained is plotted to the same axes as the graph which gives 
the ideal or prophetic curve of healing. Two such ideal 
curves and also the actual observed curves are represented in 
our diagrams. 

When the observed area is found markedly greater than 
that determined by the ideal curve, the indication is that 
there is still infection in the wound. This is the case de- 
picted, as will be noted, in the smaller diagram. A rather 
surprising and unexplained situation occurs frequently when 
the wound-surface heals more rapidly than the ideal curve 
would indicate ; in this event secondary ulcers develop which 
bring the curve back to normal. This is the type which is 
represented by our larger diagram. 

This application of mathematics to medicine is largely due 
to Dr. Alexis Carrel of the Rockefeller Institute of Medical 
Research. He noted that the larger the wound-surface, the 
more rapidly it healed, and that the rate of healing seemed to 
be proportional to the area. This proportionality constant is 



LAWS OF GROWTH . 431 

not the same for all values of the surface or we would have 
an equation of the form, 

in which -Si is the area at the time that the wound is rendered 
sterile and observations to be plotted really begin. 

The actual formulas, as developed by Dr. P. Lecomte du 
Xotiy of Base Hospital 21, Compiegne, France, are 

{) 6S, ' 

giving the characteristic constant of the wound. 

Si is the measure of the area, first observation ; S 2 is a 
second measurement taken after 4 days. 

(2) S, = S_ 1 [l-f(4 + vTn)], 

wherein S n is the area after 4 n days ; similarly, S n ^ is the area 
after 4(?i — 1) days, etc. ; each ordinate is obtained from the 
preceding ; i is the constant as determined above. 

Recent experiments by Dr. du Noiiy show that there is a 
normal value of i dependent upon the age of the individual 
and the size of the wound, and that the individual index as 
determined by two observations will doubtless reveal facts 
concerning the general health of the individual. 

The data given are taken from the Journal of Experimental 
Medicine, reprints kindly furnished by Major George A. 
Stewart of the Rockefeller Institute. The diagrams are 
reproduced from the issue of Feb. 1, 1918, pp. 171 and 172, 
article by Dr. T. Turner and R. Desmarres, Auxiliary Hospital 
75, Paris. 

6. Damped vibrations. — The combination by multiplication 
of ordinates in the two functions, y = e~*i' and y = sin k 2 t, which 
we have seen to be fundamental in the mathematical interpre- 
tation of many phenomena of nature, gives a formula which 
also has wide application. 

The formula 

y = &-H sin k 2 t, 



432 



UNIFIED MATHEMATICS 



expresses the law by which the decrease in intensity of the 
vibrations defined by y = sin k 2 t may be determined, under 
certain conditions. 




Damped vibration curve by multiplication of ordinates 

Thus a pendulum swinging in the air, when the friction is 
proportional to the velocity, has this form of equation as the 
equation of motion. We have indicated on our diagram the 

curves 

y=sinf|+2-*f\ 

V = er* 

and y = — e~ ( 
and the damped vi- 
bration curve 
y = e~ l sin 2 wt. 
The student should 
check the values, re- 
drawing all the curves 
on double the scale of 
the illustration in the 
text. 

A beautiful damped vibration curve is obtained by the 
discharge of an electrical condenser. In our illustration the 
equation 

y - e~l l sin/2 irt + ^ j 




Damped vibration produced electrically by the 
discharge of a condenser 



LAWS OF GROWTH 433 

represents quite closely the curve, using the maximum ordi- 
nate as unity, i.e. when t = 0, and on the horizontal axis the 
measure of the time in seconds of one complete vibration is 
taken as unity ; in the electrical occurrence represented by 
this photograph, and the corresponding light phenomenon 
which produced the photograph, the action took place in about 
4L of one second, and -^to °^ one secon( i is approximately the 
time of one vibration on this curve. 



PROBLEMS 

1. Plot the following curves, in the order given: 

y = sin 2 irt r , 
taking two half-inches to represent t = 1 on the horizontal 
axis and 6.2 half-inches to represent unity on the vertical axis ; 
use the graphical method. 

V = e~\ 
taking 6.2 half-inches to represent unity on the vertical axis, 
and two half-inches to represent one second on the horizontal 

axls * 2/ = e"^sin27r^, 

by multiplication of ordinates. 

Note that by taking five half-inches to represent unity on the vertical 
axis each half -inch represents .2 and each twentieth of an inch represents 
.02. These facts are to be used when you multiply ordinates to obtain the 
third of these curves. For the values of the powers of e consult the table 
at the back of the book. 

2. Given that an automobile wheel which is revolving freely 

at the rate of 400 revolutions per minute is allowed to come to 

rest by the action of the friction and air resistance ; assuming 

that the subsequent velocities per minute are given at the end 

of t minutes by the equation 

_t_ 
<u = 400e 10, 

to determine these velocities, plot the graph of the func- 
tion. At what time will the number of revolutions be 



434 UNIFIED MATHEMATICS 

reduced to approximately 200 per minute ? to 100 ? to 50 ? 
to 3? It may be assumed that at 3 revolutions per minute 
the law no longer holds and that the wheel will stop at about 
that time. 

3. Given that the horizontal displacement of a second 
pendulum is 4 inches, and that the horizontal displacement is 
given by the equation 

x = 4 cos 2 wt, 

and that, the amplitudes are decreased according to the " law 

of organic decay," the position' being given at any instant by 

the equation A -■*- . 

^ x — 4 e ioo cos 2 -n-t. 

Find the displacement of the pendulum after 10 seconds ; after 
100 seconds ; after one hour. When does the pendulum have 
a displacement of only 1 inch ? of J inch ? of T L inch ? 

This type of retardation is found when the friction is 
proportional to the velocity. 

4. Given that a fly-wheel revolving freely with a velocity of 
500 revolutions per minute is allowed to come to rest. If 
the velocity at the end of t seconds is given by the equation 

v = 500e io, 
find the velocity at the end of 10 seconds ; at the end of 100 
seconds ; at the end of 30 seconds. When will the velocity be 
reduced to 1 revolution per minute ? 

With a heavy oil as lubricator heavy fly-wheels follow 
approximately this law. 



CHAPTER XXVIII 

POLAR COORDINATES 

(See Section 3, Chapter VII) 

1. Uses. — Tor many purposes the representation of func- 
tions by the system of polar coordinates is desirable. Thus, 
effective pressure on the crank head by the piston head varies 
for every angle. It is convenient to give this pressure- 
diagram in polar coordinates ; on every radius is plotted a 
length representing graphically the effective turning pressure 
on the crank for that angle. 

2. Plotting in polar coordinates. — The coordinates of points 
which satisfy an equation given in polar coordinates are 
obtained precisely as in rectangular coordinates. An equation 
in polar coordinates involves r and 0, radius vector and 
vectorial angle ; by substituting in the given equation particular 
values of one of the variables and solving for the correspond- 
ing values of the other points on the curve are obtained. 

Illustrative problem. — Plot the curve 

r = 10 sin 2 6. 

Note that when 6 = 10°, r = 10 sin 20° = 3.42, which 
length is plotted on the 10° line. Complete the work, show- 
ing how the second loop and other loops are obtained, by- 
giving to 6 values increasing by 5° intervals np to 360°. Note 
that no further computation is needed. Follow the progress 
of the curve on the diagram given on the next page. 
435 



e 


r 


0° 





10° 


3.42 


15- 


5.00 


20° 


6.43 


25° 


7.66 


30° 


8.66 


35° 


0.40 


4CP 


9.85 


45° 


10 


W° 


9.85 



436 



UNIFIED MATHEMATICS 




The formulas for transformation from rectangular coordi- 
nates to polar coordinates should be noted : 

x = r cos 0, 
y = r sin 6. 

The formulas for transformation from polar to rectangular 

coordinates are : . 

r = V # 2 + y'^ 



Y\ 










::::::::::p:(£0-): 




::::::::;53ffiiftt 




g:^f:=:ME: 


: ~ ?- 


"' -- - ~f?T 


^^-"j^ 


IliEIEEEIlilEIEg 


^EEEEEEEEEEE 


:::::::::::::::£ 



= arctan ?. 
a; 



and sin # = 



or 



cos 



V# 2 + 2/ 2 



Vx 2 + y 2 



POLAR COORDINATES 437 

PROBLEMS 

1. Plot the curve r = 10 sin 0. Prove that this is a circle. 
Take any point on the circle, as (r, 0), and show that the co- 
ordinates satisfy the equation. 

2. Plot the following curves : 







(a) 


r sin = 


= 5. 






(P) 


r cos - 


= 10. 






(«) 


r sin (0 


-30°) =10. 






(d) 


r = 10 sin 3 0. 






(«) 


r = 10.- 


- 10 cos 0. 






(/) 


r==5. 




3. Plot 


the 


curve 


r = 2 a 


tan sec 


angular coordinates. 


This curve is a "c: 



transform to rec- 
cissoid" and can be 
used in the " duplication of the cube " problem. 

4. Plot the curve r = 10 sec 6 + 5. This is a " conchoid of 
Xicomedes," and can be used to effect the solution of the 
problem to trisect any angle. 

5. Plot the polar diagram of effective pressures on the 
crank for different angles of ; use the data of problem 2 in 
the problems given under piston-rod motion. 

6. Plot the curve r — 10 — 10 cos 0. This curve is called 
a " cardioid " because of its shape. 

7. Plot the curve r = 10 — 5 cos 0. This is called a 
" limaqon of Pascal." 

8. Plot r = 10 — 20 cos 6, another type of limaQon. 

9. Show that the polar equation of any conic is 

2m 
1 — e cos 6 
wherein 2 m is one half of the right focal chord. 

10. Plot the parabola 

10 
r= . 

1 — cos $ 



438 UNIFIED MATHEMATICS 

11. Plot the hyperbola 



10 
r 



1 - 2 cos 

For what values of 6 is r infinite in value "? What directions 
do these values give ? Are these lines from the origin then 
the asymptotes ? 

12. Plot the spiral of Archimedes given by 

r = 10 0. 

13. Plot the hyperbolic spiral given by 

r = — 

e' 

14. Plot r= sin 2(9. 

15. Plot r = sin -f- sin 2 6, and compare the polar with the 
Cartesian (aj, y) representation. 



CHAPTER XXIX 
COMPLEX NUMBERS 

1. Object. — In the study of the number field, indicated in 
our first chapter, we found that in the extraction of square roots 
we were limited to positive numbers. Again in solving quad- 
ratic equations, and in the discussion of the roots of algebraic 
equations, Ave found that no number of the kind we had con- 
sidered could occur as the even root of a negative quantity. 
We can extend the number field, removing the limitation that 
square roots and even roots must be taken of positive quan- 
tities only, by creating another class of numbers, complex 
numbers. These numbers, after the fundamental operations 
with them have been properly defined, apply to our algebraic 
equations, x and the constants being complex numbers. In 
the- extended number field it is possible to prove that every 
rational integral equation has a root and that such an equation 
of the nth. degree has n roots. 

2. Complex numbers. — We define V — 1, designated by ?', as 
a number which, multiplied by itself, equals — 1 ; this re- 
quires, then, an extension of the meaning of multiplication 
and a reexamination of the fundamental processes as applied 
to the old numbers with this newly found number and other 
new numbers which follow directly from it. This discussion 
is given graphically in the next section. The square root 
of any other negative number, — a, is regarded as VaV — 1, 
or Va • i. Such a number, e.g. V— 7, is called a pure imag- 
inary. To add a pure imaginary to a real number both must 
be written and the combination is called a complex number. 

439 



440 



UNIFIED MATHEMATICS 



Thus, x 4- yi and a -+- bi, or — 3 + 2V— 1 and V5 — V3 i, 
are complex numbers. 

Addition and multiplication are explained graphically in 
sections 6 and 7 below. 



-- 


T 


















*■* ^ N X 


4 \X 


-; I ,11 ±_L_ 




- 1 ur 
















■\ 







• 3. Graphical representation. — Our real numbers can all be 
conceived graphically, as well as analytically, as derived from 
the unit 1. Thus, integers are obtained by the repetition of 
the unit ; fractions are obtained by 
the subdivision of the unit ; and nega- 
tive numbers are obtained from the 
negative unit, which in turn is ob- 
tained by reversing the direction of 
the positive unit. Analytically the 
imaginary unit repeated as a factor 
gives — 1 ; graphically then we would 
desire an operation which repeated 
gives a reversal of direction. How 
is the reversing from +1 to — 1 
effected? Evidently by turning the 
an angle of 180° or -180°. The 
V — 1, or i, "can be regarded then as represented by the 
middle position of this rotating unit, and the upper position is 
regarded as -+- & and the lower as — i. This vertical line is 
taken as the axis of pure imaginaries. Thus, V — 4, or 2 i 9 
is represented two units up on this axis and — V — 2 is repre- 
sented V2 units down on this axis. 

A complex number, x-\- yi, may now be uniquely represented 
by the point (x, y) in the complex plane, in which the y-axis 
coincides with the vertical axis of pure imaginaries. 

The fundamentally important facts concerning these num- 
bers are : 

1. Complex numbers are combined according to the laws cfthe 
real numbers (which we have discussed in the first chapter), 
noting that i 2 = — 1. 



The imaginary unit ob- 
tained graphically 

positive unit through 



COMPLEX NUMBERS 



441 



2. The combination of any tivo or more complex numbers, by 
the operations of addition, subtraction, multiplication, division 
(except by zero), involution, and evolution (with certain excep- 
tions), always produces a complex number. 



dK 



-2,+,V3i 



ffi 



£ 



jm 






2V2H: 



iY 



Representation of three complex numbers 

Two complex nnmbers of the form x ■+- yi and x — yi, sym- 
metrically placed with respect to the axis of reals, are called 
conjugate complex nnmbers. Their sum and their product are 
real numbers. 

4. Complex roots in pairs. — In any rational integral algebraic 
equation with real coefficients, if a + bi is a root of the equa- 
tion, then a — bi is also a root of the equation. The proof 
depends upon the fact that when a + bi is substituted in 

a x n -f- ciiX"' 1 + a 2 x n ~ 2 -f- "• a n , 

the resulting expression is of the form P + Qi, in which P and 
Q being real numbers, P involves powers of a and the even 
powers of bi, and Q is obtained from expressions involving 
odd powers of bi. Now if 

P+Qi = 0, 

then P—0 and Q — 0-, otherwise you have a real number 
equal to a pure imaginary. Substituting a — bi for x in 



a x n -j- a x x n ~ l -j- 



442 



UNIFIED MATHEMATICS 



changes the signs of the terms involving the odd powers of i, 
and does not change the sign of the even powers. Hence 
a — bi substituted gives 

but P = and Q = 0, hence P— Qi = 0. Therefore a — bi is 
also a root of the equation if a + bi is a root. Complex roots 
go in pairs. 

Illustrative problem. — 1. Find the product of 2+V — 3 by 
3 — V — 2 and put the product in the x + yi form. 



:=EE::=i=EEE=::E::=EEE:::E:::E:E::Ei=TITS=5g==: 


:::::::^:::::::4::::::=:=:::::=:::::-3f^^:(sp^4^$ i= = : 


i "EE u' '** 


EEEEEEE|^jg|^^|^:!:::EEEEEEEEEE:EEEEEEEEE 


[||mM| P 


eeeeeeeee~|||eeeeeeeeeeeeeeeeeeeeeeeeeeeJeeeeeeeeeee 


eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeIeeeeeeeeeeee 



Graphical representation of the product of 2 + V3 i" by 3 — a/2 i 

(2 + V3 j)(3 - V2 i)= 6 + 3V3"i - 2a/2 i - V6 i 2 , 
but i 2 = — 1 , giving as product, 

6 + V6+(3V3 -2y/2)i. Ans. 

6 +V6 = a: 3V3 -2V2= b. 



2. Divide 3-fV— 3 by 5 — 2 V — 3 and express the 
quotient in x -f- 2/*' form. 

3 +\ /zr 3 _ 3-f V3£ _ (3+ a/3 Q(5 + 2V3Q _ 15+ 6P + 11VSJ 
5_2a/^3 5-2V3£ (5-2V30(5 + 2V30 
_ 9 + llv3 4 _ _9_ llv/3i 
37 ~37 37 



25 - 12 i 2 



3. Factor x 2 -f- ?/ 2 into complex factors, linear in x and y. 
x 2 + y 2 =(x + fy) (a- _ jy). 



COMPLEX NUMBERS 443 

PROBLEMS 

1. Write tlie conjugate complex numbers : 

a . 3 + V-~2. d. 1 + i. 

b. -A-2L e. -V-7. 

c. _3_V2-V^2. / & 

2. Eationalize the denominator in the following expressions 
by using the conjugate complex number as multiplier, reduc- 
ing the quotient obtained in this way to the form a 4 hi. 

5_V^2 , 2 



a. 



3+V-2 1 + i 

5 V5 

e. 



-4-2* _ V ^7 

c> 3+V2+2V-2_ 5-V-3 

_3_V2-V-2 J ' 3 

3. Write the following expressions in the form a 4- hi : 
a. £ 4 i 2 + ** 3 + *' 4 . „ 2 2 



6. t 5 4 3 ?: 6 + 3 V + 4 i 8 . 3 4 2 i 3 - 2 i 

c. i 10 + i 20 . 2 2 

d. 2< + 3* + 4*. /• 3^2^-3321' 

4. Locate the points represented by the complex numbers 
in problem 1. 

5. Square — - 4- — - i ; square — — i. These are roots 

of a? - 1 = 0. Multiply - i + ^?i by - 1 - ^i. What is 

Z Z Z Z 

the cube of -- + ^-i? 

2 2 

6. Square ~ and \_' . Give an equation with real 

V'2 V2 

coefficients which these numbers satisfy. What are the 
square roots of i ? 



444 



UNIFIED MATHEMATICS 



5. Vectors. Polar representation of complex numbers. — The 

complex number x + yi may be regarded either as determined 
by the point P(x, y) or by the vector OP, which by its length 
and direction determines the position of P. The angle which 
OP as a ray makes with the cc-axis is called 6, the amplitude or 
angle, and the length OP is called r, the modulus of the com- 
plex number. In other words, the polar coordinates of P are 

(r,0); 




cos 



= -, and sin0=:^. 



The complex number may be writ- 
ten in the form 

r(cos 8 + i sin 6), 

which is termed the polar form. 



Modulus and amplitude of 
a complex number 



The modulus, r or ^/x 2 + y 2 , is a 

positive number representing the 

length of the vector or the distance of the point x + iy from 

the origin. This modulus is sometimes called the stretching 

factor or the tensor ; see section 7. 



6. Addition of vectors. — When the complex number is repre- 
sented by a vector, the sum of two complex numbers will be 
represented by the diagonal of the parallelogram formed by 
the two vectors ; see Chapter IX, section 2. The student 
should verify the fact by a diagram. 



7. Product of complex numbers. — Given two complex num- 
bers, either in polar form or in rectangular form, the product 
of the two numbers is also a complex number ; further the 
modulus of the product is the product of the moduli, and the 
amplitude or angle of the product is the sum of the amplitudes 
of the factors. 



COMPLEX NUMBERS 445 



Let 



V\ 



rx(cos di + i sin X ) ; X\ + s^i, mod V#i 2 + ?/x 2 , ainpl X = tan x 

and 

?- 2 (cos 2 -f i sin 2 ) ; #2 + 2/2*? m °d V# 2 2 + 2/ 2 2 > ampl 2 = tan -1 •— 

be two complex numbers. Their product is 

^[cos di cos 2 — sin 9 l sin 2 ■+- i(sin 0! cos 2 + cos 0, sin 2 )] 

«i*2 - 3/1^2 + f'(-M2 + «^i), m o d Va?i 2 JB2 2 + 2/iV + ^i 2 2/2 2 + ^V 

amplg 8 = tan- i a?1 - V2 + a;2yi - 

'^1^2 — ,W2 

The polar product may be written 

r,r,[ooB (0! + 2 ) + i sin (0 X + 2 )], 
showing that the product of the moduli r x and r 2 is the modulus 
?V' 2 of the product and the amplitude is. X + 2 , the sum of the 
amplitudes. It is left as an exercise for the student to show 
that the analytical expressions for modulus and amplitude 
establish the same facts. 

"When any complex number is used as a multiplier, the 
modulus of the product is the modulus of the multiplicand 
stretched in the ratio of the modulus of the multiplier to 
unity. For this reason the modulus is sometimes termed the 
stretching factor. 

8. De Moivre's theorem. — Evidently, if X and 2 are set 
equal to 0, our product formula may be written : 

(1) [r(cos + i sin 0)] 2 = r 2 (cos 2 + i sin 2 0). 
Evidently by mathematical induction, by simple introduction 
of one further factor r(cos -\- i sin 0) at a time, it can be 
shown that 

(2) [r(cos + i sin 0)] n = r n (cos.w0 -f- i sin nO). 

This theorem, which holds for all values of n, is called De 
Moivre's theorem. We have proved it only for n an integer. 
Taking the wth root of each member of equation (2), we have 

(7") "(cos nO + i sin »0) n = '/'(cos -f- i sin 0). 



446 UNIFIED MATHEMATICS 

Let r n = k and n6 = 0', which, as no limitation was imposed on 
0, imposes no limitation as to value on 0', and we have 

[&(cos 6' + i sin 6')~] n = k"( cos — \- i sin — ), 

\ n 71 J 

i.e. our formula holds for a fractional exponent of the form -. 

n 

By raising to the mth power both sides, it can be shown to 

hold for any fractional exponent. 

For n = - 1, [r(cos + i sin 6>)] _1 - 



r(cos + i sin 6) 
cos — i sin 



r(cos 2 + sin 2 6) 
= r -1 (cos 6 — i sin 0), 
whence 

[>(cos + i sin 0)]" 1 = r-^cos (- 6) + i sin (- (9)], 

which establishes the formula when n = — 1. By raising both 
sides to the >ith power, n any rational number, the theorem is 
established for all rational exponents. 

The theorem can be established also for irrational values 
of n. 

PROBLEMS 

1. Write the following complex numbers and their conju- 
gates in polar form, giving modulus and amplitude : 

a. =± + &l /" ~ V3T: 

2 2 g- i. 

c. _ 3 -V2-V^2. 7i ' 23 - 

d. 1 + 1. . 1 V3 . 

e. 3 + V^2. '" 2 2 *' 

2. Show that (cos 30° + i sin 30°) 2 = cos 60° + i sin 60°, by 
multiplication. 

3. Show that (cos 30° + i sin30°) 3 = i, i.e. cos 90° + i sin 90°. 

4. Show that = cos ( - 30°) + i sin ( - 30)°. 

cos 30° + i sin 30° 



COMPLEX NUMBERS 447 

5. Show that (cos 60° + i sin 60°)^ = cos 30° + i sin 30°. 

6. What is the value of (cos 30° + i sin 30°)^ by De Moivre's 
theorem ? 

7. Plot, nsing 2 inches as 1 unit, cos 15° + i sin 15°, 
cos 30° + i sin 30°, cos 45° + i sin 45°. 

n ™ , ,, ■ , -r, 360° , . . 360° , , 

8. Plot the point B, cos h i sin ; connect by a 

chord with the point A, cos 0° + i sin 0° ; take this length as a 
chord, successively seven times on the unit circle about O. This 
chord is the side of a regular inscribed polygon of seven sides. 

9. Roots of unity. — Plotting the solutions of the following 
equations on the complex number diagram, 

(See Section 3, above), 

x —1 = gives one point, 1 ; 

x 2 — 1 = gives two points, 1 and — 1 ; 

.x* 3 — 1 = gives three points, 1, — - -\ i, i ; 

x 4 — 1 = gives four points, 1, — 1, i and — i ; 

X* — 1 = gives six points, 

. , 1.V3. 1 V3.1.V3. , , 1 V3. 
' 2 2 2 222 2 2 

X s — 1 = 0, or (x 4 — l){x 4 + 1) = 0, gives eight points, which 
may be obtained by methods of quadratic equations. For 

x 4 + l = x 4 + 2x 2 + l-2x* = (x 2 +.1)2 -(V2 x) 2 
= (z 2 + 1 - V2 a?) (z 2 + 1 + V2 aj), 

whence, ^-V2x + 1 = 0, x= V2±V - 2 = ^g±^gi 

and ^ + V2» + l = 0,»=- v ^ ±V ^ = -^g ± ^gi. 



448 



UNIFIED MATHEMATICS 



1111 

^—1 = gives eight points, 1, — 1, i, — i, — - _i_ — - i, — — 

V2 V2 V2 V2 

1 1 . , 1 1 . 

H r i, and 1. 

V2 V2 V2 V2 

a? 5 — 1 = gives five points ; see the solution obtained in 
problem 4, page 97. The solutions are 



i V5 + 1 V10-2V5. V5 + 1 V10-2V5 . 

' ~A i l > A 1 A h 



1 + V5 , V10 + 2V5, nrxA -1+V5 VlO + 2V5; 
4 



-\ — — '-^-^ i, and - 



Plotting these points on the complex diagram gives the ver- 
tices of a regular pentagon. 

Graphically representing these points we have the following 
diagrams : 




Graphical solutions of x 3 — 1 = 0, x 1 — 1 = 0, x 6 — 1 

and x 5 - 1 = 



= 0, x 12 - 1 = 0, x 8 - 1 = 0, 



COMPLEX NUMBERS 449 

The roots of any equation of the form 
x n — 1 = 0, n integral, 

are, aside from + 1 or — 1, complex numbers. Since any real 
number less than 1 when multiplied by itself gives a number 
less than 1 and since any real number greater than 1 multiplied 
by itself gives a number greater than 1, it follows that the 
roots other than 1 or — 1 are complex. Further, the modulus 
of each of these complex roots is a real number which, taken n 
times as a factor, produces 1 ; hence the modulus of any nth 
root of unity is 1. We need then to know only the real part of 
any root of unity to plot it, since the root itself, having a 
modulus 1, lies on the unit circle, x 2 + y 2 = 1. 

The nth roots of unity can be obtained graphically by rinding 
the angles which repeated n times give 360° or integral 
multiples of 360°. 

Thus for the twelfth roots of unity these angles are 0°, 30°, 
60°, 90°, 120°, 150°, 180°, 210°, 240°, 270°, 300°, and 330°. 
Writing the corresponding complex numbers in polar form we 
have the twelve twelfth roots of unity. If we went farther, 
taking 360°, 390°, 420°, ••• , we would simply repeat values 
already obtained. The twelfth roots of unity are, then, 

' 2 + 2 '2 .2 \' 2 + X ' 2 '2 ' ' 

V3 1 . 1 V3 . . , 1 V3 . A , V3 1 . 
1, -^—h —h ^ i, and + — 1, 

2 2 ' 2 2' '2 2' 2 2 ' 

as on the figure. 

10. Historical note. — Just as negative numbers were gener- 
ally accepted only after a graphical scheme of representation 
of these numbers was introduced by Descartes, so imaginary 
numbers were neglected and even rejected by mathematicians 
until a graphical system of representation was found. 

In 1797 a Norwegian, Caspar Wessel, presented the scheme 
of representation of complex numbers to the Danish Academy, 
but public recognition of his work is only recent. A French- 



450 UNIFIED MATHEMATICS 

man, J. E. Argand, discovered the same system independently 
in 1806 and it appears that the German J. C. F. Gauss in 1831 
again independently rediscovered this graphical method. The 
latter made extensive use of the diagram and since then these 
numbers play a vital role in the development of algebra. 
Quite recently the practical importance of these numbers has 
been more generally recognized by physicists and engineers. 
Applications have been made to problems in electricity, to the 
steam turbine by Steinmetz, and to numerous other vector 
problems. 

The term imaginary is a misnomer, as our development 
shows. So far as actuality is concerned, 3 + V— 3 exists as a 
number quite as much as 3 or V3 ; all numbers are the product 
of intelligence reacting on the experiences of life, and in 
this sense all numbers are imaginary, the product of the imagi- 
nation. 

11. Mathematical unity. — The complex numbers are fittingly 
chosen to conclude our treatment of plane analytic geometry, 
elementary algebra, and elementary trigonometry since, as the 
observant student will have noticed, we have here involved 
the fundamental principles of these subjects as well as theo- 
rems of plane geometry. 

We might note that while regular polygons of seven and 
nine sides cannot be constructed with ruler and compass, since 
the solutions of these equations lead to cubics which cannot be 
solved in terms of quadratic irrationalities, there are other 
polygons having a prime number of sides which can be so con- 
structed. Gauss, when only 19 years old, showed that the 
polygon of 17 sides, and, in general, the polygon of sides 
2 2 " + l in number, when this number is prime, can be con- 
structed with ruler and compass. The corresponding alge- 
braic fact is that x m — 1 = 0, when m is a prime number equal 
to 2 2 " -f- 1, is solvable by the methods of quadratics, and the 
roots can be expressed in functions involving only square 
roots. 



COMPLEX NUMBERS 451 

PROBLEMS 

1. Solve .x* 3 — 1 = and plot the points on the polar diagram. 

2. Solve similarly z? + 1 = and plot. 

3. Solve x h 4- 1 = 0. Using the results given for x b — 1 = 0, 
write all the solutions of x 10 — 1 = 0. 

4. Derive the formulas for cos 2 and sin 2 0, using De 
Moivre's theorem. 

Hixt. cos 2 6 + i sin 2 6 = (cos -f i sin 0) 2 . Square the right-hand 
member and then eqaate cos 2 6 to the real part and put sin 2 equal to 
the coefficient of i. 

5. Derive formulas for cos SO and sin 3 by the process 
of problem 4. 

6. Show on the diagram how to obtain the twelfth roots of 
unity. What equation do these numbers satisfy ? 

7. Show geometrically and algebraically how you can obtain 
the solutions of the equation 

£24 _ 1 = 

from the complex diagram. Use also the formulas for sin \ 
and cos \ to obtain sin 15° and cos 15° from sin 30° and 
cos 30°. 

8. Solve x 8 — 1 = 0, and plot the points on the complex 
diagram. 

9. Solve x 16 — 1 = 0. What angles are involved ? 



CHAPTER XXX 

SOLID ANALYTIC GEOMETRY: POINTS AND LINES 

1. The third and fourth dimensions. — We have found that on 
a line the position of any point may be given by a single 
number, x, which locates the point with reference to one fixed 
point "on the line. The single number, commonly x, represents 
distance, in terms of some unit of length, from the point of 
reference, and direction by means of a + or — sign. In a 
plane the position of any point may be given by a pair of 
numbers which locate the point with reference to two fixed 
lines in the plane. The two numbers, x and y commonly, 
represent the distances in determined order, in terms of some 
unit of length, from each of the two given lines of reference, 
and direction as before. By analogy, continuing with the 
proper changes, it 'is obvious that in space the position of any 
point may be given by a set of three numbers which locate the 
point with reference to three fixed planes in space. The three 
numbers, x, y, and z commonly, represent the distances in 
determined order, in terms of some unit of length, from the 
three given planes of reference, and the direction in each case 
is determined by the algebraic sign of the number. If the 
analogy could be continued we could state that the position 
of any point in a four-dimensional space would probably be 
given by a set of four numbers which locate the point with 
reference to four given " three-dimensional " spaces. The four 
numbers, x, y, z, and iv commonly, would then represent the 
" distances " in determined order, in terms of some unit of 
length, from each of the spaces of reference. Without a pre- 
cise definition of what we mean by " distance " of a point in 

452 



SOLID GEOMETRY: POINTS AND LINES 453 

" four-dimensional space " from a " three-dimensional space " 
these analogies must be regarded as purely fanciful, and devoid 
of physical significance. 

Location of a Point 



Upon or in a 


With reference, to 


By means of 


line, 


one point, 


one variable, 


one dimensional. 


zero dimensional. 


X. 


plane, 


two lines, 


two variables, 


two dimensional 


one dimensional. 


0, y). 


space (ordinary), 


three planes, 


three variables, 


three dimensional. 


two dimensional. 


(x, y, z). 


hyperspace, 


four three-spaces, 


four variables, 


four dimensional. 


three dimensional. 


(X, 2/, Z, w). 


n-space, 


n(n — l)-spaces, 


n variables, 


n dimensional. 


(n — 1) dimensional. 


(Xi, x 2 , x 3 , ••• x n ). 



2. Space coordinates. — The position of a point in ordinary 
space is determined by location with respect to three inter- 
secting planes, called the coordi- 
nate planes. Just as our lines of 
reference were chosen perpen- 
dicular to each other, for conven- 
ience, in plane analytics, so here 
the planes of reference are taken 
mutually perpendicular, like the 
three sides of a box or like the 
front wall, the floor, and the left- 
hand wall of a room. The three 
lines of intersection of these 
planes with each other in pairs 
are called the axes of coordinates, 
designated as x-axis, y<ixis, and z-axis; the three planes are 
named xy-, xz-, and yz-planes respectively ; the point common 
to the three planes and to the axes is called the origin. 







j 














I 










































































•J?i+T 




_J . 1 










— —-, 








































































: — 






' ' ' ' 




■ 








1 :» 










■ i JC 




















--/-&- 










/ / " 






; '■ '■■ ■ ' - 












~ 7 










r 










? 










-U 








MM 


1 



Axes in space 



454 



UNIFIED MATHEMATICS 




The numbers x, y, and z represent respectively distances 
from the yz-, xz-, and .T?/-planes ; direction is determined as 
indicated by the arrowheads upon the diagram. No general 
agreement has been reached as to which axis to use as the 
vertical axis. The system shown is called a right-handed 

system, since the 90° rotation 
of the positive ray of the 
#-axis into the positive y ray 
advances a right-handed screw 
along the z-axis, and similarly 
with the other axes, by cyclical 
interchange in x, y, z order. 
The positive directions of 
these three axes can be repre- 
sented by the thumb, first 
finger, and second finger of 
the right hand. 

To determine the coordi- 
nates of any point P in space, 
planes are drawn or conceived 
through the point parallel to the coordinate planes ; the dis- 
tances OL, OM, and ON cut off on the axes are given with 
proper sign as the coordinates (x, y, z) of the point P. 

Space is divided by the coordinate planes into eight divi- 
sions, called octants. The signs of the coordinates of any point 
within an octant are given in xyz order to distinguish the 
octants. Thus the + — — octant is at the right, below, and 
back. 

To every point in space corresponds one set of coordinates 
and only one, and, conversely, to every set of three numbers 
corresponds one and only one point in space. When a point 
is given by its coordinates, the position is determined on the 
diagram by passing a plane through the <c-axis at the x of the 
point, parallel to the yz-plane ; on the intersection of this 
plane with the ^-plane indicate the y coordinate. The third 
coordinate must be represented in perspective, and the direc- 



O(0, 0, 0); £(3.5, 0, 0); Af(0, 3, 0); 

N(0, 0, 2); 0(3.5, 0, 2); /?(3.5, 8, 0); 

S(0,3,2); P(3.5,3,2) 



SOLID GEOMETRY: POINTS AND LINES 455 

tion and length of these units in perspective are taken paral- 
lel and equal to the units represented on the third axis. 




Drawing by L. Makielski. 
An artist's conception of a rectangular solid in space 



3. Fundamental propositions of solid geometry. — The follow- 
ing propositions of solid geometry have constant application 
in our further work. The student would do well to review 
these propositions in any elementary work on solid geometry 
and further to verify the reasonableness of these propositions 
on our figures. 

a. Two planes intersect in a straight line. 

b. If a line is perpendicular to each of two intersecting 
lines, it is perpendicular to the plane of the two lines, i.e. it 
is perpendicular to every line in the plane of the two given 
lines. 

Thus, PQ on the diagram below is perpendicular to QL, to 
QM, to QD, and to every line in the az-plane which passes 
through Q. A line in the a;z-plane which does not pass through 
Q does not intersect PQ, but the angle which it makes with 
PQ is defined as the angle which any parallel to it which does 
intersect PQ makes with PQ. This will then be a right angle. 

c. The angles between two pairs of parallel lines are equal 
or supplementary. 

d. If two planes are perpendicular to a third, their inter- 
section line is perpendicular to the third. 



456 



UNIFIED MATHEMATICS 



e. The dihedral angle between two planes is measured by 
the plane angle formed by two lines, one in each plane, both 
perpendicular to the edge of the dihedral angle. 

/. If a line (PQ) is perpen- 




The dihedral angle, between two 

planes, is measured by the 

angle between two lines 



dicular to a plane (xz) and from the foot of the perpendicular 
a second perpendicular (QL or QD) is drawn to any line 
(OX or NL) in the plane (zx), then the line connecting any 
point (P) on the first perpendicular to the intersection point 
(L or D) is perpendicular to the line (OX or NL) in the plane. 

Note. — We will refer to these propositions as 3 a, 3 &, 3 c, 3d, 3 e, 

and 3/. 



4. Vectorial repre- 
sentation. — For 

some purposes it is 
convenient to think 
of three numbers (x, 
y, z) as representing 
the vector from the 
origin to the point. 
The length OP of 
the vector is called 
r, and the vectorial 





i 7 
















__ ,:0-2$$jS' 


y'^- 


-,* -trL 


± x „? gar 




_ _ '-"-''& Jtt % - 


-H-*>?K -^ 








() tiJ_ iJ__:li2Lt__. ir..-.t* t- 




C S** S- 


l ?\ ^ ± ^H 




2?E *' s -«. -- -<ss- 


i. c s~ iv^v- 






/ \ 










z 





Point P(x, y, z) or P(r, a, p, y) 



SOLID GEOMETRY: POINTS AND LINES 457 

angles which this vector makes with the x-, y-, and z-axes are 
termed direction angles and are represented by the letters, 
a, ft, and y, respectively. 

By proposition 3/, the triangles PLO, PQO, and PNO are 
right triangles. Hence 

x = r cos a, y = r cos (3, z = r cos y. 



Further, the square on the diagonal OP( = r) of our rectangu- 
r pris 
whence 



lar prism is the sum of OQ 2 and QP 2 , and OQ 2 = ON 2 + NQ\ 



f- = r 2 cos 2 a + r 2 cos 2 (3 + r 2 cos 2 y, 

or cos 2 a + cos 2 (3 + cos 2 y = 1. 

Evidently, also, 

r 2 = a 2 -f y 2 + z 2 . 

5. Parametric equations of a line. — The equations 

x = r cos a, ?/ = r cos /?, 2 = r cos y, 

when «, (3, and y are fixed and r is a variable parameter, serve 
as the equations of the straight line OP. Cos a, cos (3, and 
cos y are called direction cosines of this line. 

Evidently any point E(a, b, c) on the line OP produced in 
either direction satisfies this relationship, if r is taken as the 
distance from to E. If E is on the other side of the origin 
from P, then the direction angles of the vector OE are supple- 
mentary to a, f3, and y and have cosines opposite in sign to 
cos a, cos f3, cos y. In this case r is taken as negative and it is 
evident that with this interpretation the coordinates of the 
point E satisfy the given equations. 

A line which does not pass through has the same direction 
cosines as the line parallel to it through 0, positive directions 
on both being the same. If such a line passes through 
P(%i, ?/i, Zi) in space, the corresponding parametric equations 
are 

x — Xi — r cos «, y — y x = r cos (3, z — z x = r cos y. 



458 



UNIFIED MATHEMATICS 



In plane analytics the corresponding formulas for straight 
lines are 

x = r cos a, y = r cos (3, where cos ft = sin a 
an x — a^ = r cos a, y —y l = r cos /3. 

The student should check these formulas and show their rela- 
tions to the ordinary equations used. 

6. Distance formulas and spheres. — 

r 2 = x 2 + y 2 + z 2 
is the square of the distance from (x, y, z) to (0, 0, 0). 

x 2 + y 2 + z 2 = r 2 
is an equation which is satisfied by every point on a sphere. 



sjS 



<5 



* 



1 



r/:) 



i 



The sphere : (* - /i) 2 + (y - ft) 2 -f (z - /) 2 = r 2 



r = V (x 2 - x,) 2 + (y, - ;>/i) 2 + (z 2 - z,) 2 
is the distance "between two points (a^, ?/ l5 zj and (x 2 , y 2 , z 2 ). 

(x-hy+(y-ky+.(z-iy 

represents the square of the distance from (x, ?/, z) to (h, fc, Z) 
and hence, ( <r _ jqj + ( ?/ _ k y + ( z _ qj _ r 2 

is the equation of a sphere whose center is (h, k, I) and whose 
radius is r. 



SOLID GEOMETRY: POINTS AND LINES 459 



7. Point of division. 

fc\ -\- K'2 "'1 ~r *^2 "^1 H - 1^2 



- - -- . T 


>Y T ■■ 


i i - 


7?o 








1 _ . _ ,£*_ ,' 


,S / 


± _ „5 Z ±_ 


1 -' / 


T T> * ' 






.»" >n 


1 1 Lr' / ^" ' " 


J? ,' ' r-^ 


' ' 1 | Jf -''' ' 




Hi'P, /£''" t.--' 






P,r''i| | -'' 




t^'p t- - 




-t-i X it 


±± j: yi t: t _£._ 




„.,. i._±uJ. Tl£ ;,.. I Jul£n nT 


2 - — _ — j — .^... .._--__-__.-. ..._.-.._^._.^.^ 


1 1 /I "■ " 1 _ / "'" / " / 


/ | -y. .... f 


' • ' / it v i i" - ' 


in!/ tf / 




NIL 1 " QF ■*"-** *■ ,/ 










j£ ' " Qr 


7 ! • i " 


ft\\ 


MM 



^1^3 = QlQ 3 = ^3 = X 3 - Xi = Z/3-t/l = 23- gl = ki 

P3P2 QzQi L3L2 x 2 — Xz y 2 — yz z 2 — z 3 k 2 



Step for step, and letter for letter, the proof follows that 
given for the point of division in a plane ; the only change is 
that the z-terrn is added. 



Thus, since 5^ = ^, it follows that ^- 3 = - 1 

X3./2 ^2 -^3^2 ton 



, etc. 



PROBLEMS 

1. What does the equation x — 3 or x — 3 = represent on 
a line, the z-axis, i.e. when you are considering points on a 
line ? What does this equation represent in plane analytics ? 
What does this equation represent in space analytics ? 

2. What does the equation x 2 = 9 represent in one dimen- 
sional analytics? How are these two points located with 
reference to the origin ? What does the equation x 2 + y 2 = 9 



460 UNIFIED MATHEMATICS 

represent in the xy-p\&ne ? How are the points, which lie on 
this locus, located with reference to the origin ? What does 
the equation x 2 -f y 2 + z 2 = 9 probably represent in xyz- 
coordinates ? 

3. Where do all lines lie which make an angle of + 30° 
with the x-axis ? Where do all lines lie which make an angle 
of -f- 70° with the ?/-axis ? Where does a line lie which has 
the angle a equal to 30° and the angle (3 equal to 70° ? Deter- 
mine the angle y, using the relation cos 2 a + cos 2 ft + cos 2 y = 1. 
Given that a = 30° and (3 = 60°, what is the value of y ? If 
a = 30°, (3 = 45°, what is the value of y ? 

4. Write the equations, in parametric form, of a line 
through the origin which has the direction angles a = 30°, 
/3 = 70°, and the third angle as determined in the preceding 
problem. Write the equations when these angles are 30°, 60°, 
and 90°. 

5. Write the equations of lines parallel to the two lines 
of the preceding problem and passing through the point 
(-2,3,-7). 

6. Write the equation of the sphere whose radius is 10 and 
whose center is the point ,(2, —3, 4). Find three other points 
on this sphere. 

7. Find the coordinates of the points of trisection of the 
line joining A(-2, 3, - 7) to B(2, - 3, - 4). If the line AB 
is extended through B by its own length, what are the coordi- 
nates of the point so determined ? 

8. Given that the parametric equations of a line are 

x = 3 r, y = — 2r, z = — 5r, 
find 10 points upon the line by giving to r values from — 4 to 
+ 5. Determine the direction cosines of this line. Determine 
from your trigonometric tables the angles a, j3, and y. 

9. Given the parametric equations of a line 

x — 3 = 3 r, y + 5 = — 2 r, and z — 7 = — 5 r, 
find 10 points on the locus ; determine the direction cosines 
and the angles a, j3, and y. 



ii 



2:3 



1 



==?SS 



SOLID GEOMETRY: POINTS AND LINES 461 

8. Angle between two lines. — We have had occasion to note 
that the angle between two non-intersecting, or skew, lines in 
space is defined as the angle be- 
tween two intersecting lines which 
are respectively parallel to these 
given lines. For convenience 
these parallels may be taken 
through 0. 

The angle between two lines 
having direction cosines (a lt /?i, yi) 
and (aa, /J 2 > 72) respectively is ob- 
tained as follows : The angle between two lines in 

Take any two points P l and space 

P 2 , one on each line, having vector distances r 3 and r 2 . 
Evidently 

Pj>?= r x 2 + r 9 2 - 2 r x r 2 cos 6. 
But 

7V? = (as, - x,y + (y 2 - ytf + (z 2 - z x )\ 

Equating the two values and canceling, noting that 

r x 2 = xf + y? + zf, and r 2 2 = x./ + y£ + z 2 2 , 



this gives 



Now 



Xi = r 3 cos a 1? y l = r x cos /3 b z 1 = r x cos y 1? 



and 

2- 2 = T 2 COS « 2 , 2/2 = ? '2 C0S /? 2 , ^2 = r 2 C0S 72- 

Substituting, 

cos = cos a x cos a 2 -f cos {3 X cos /? 2 + cos y l cos y 2 , 

a relation which is independent, as it must be, of the particular 
points P l and P 2 chosen. 

The corresponding formula in the plane for the angle be- 
tween two lines is 



cos 6 = cos «i cos « 2 + sill a l sin a 2 . 



462 UNIFIED MATHEMATICS 

The adaptation and the proof are left to the student as an 
exercise. Compare with formula on page 247. 

Frequently the direction cosines of a line in space are repre- 
sented by I, m, n or l 1} m 1? n ly etc. The condition that two 
lines be perpendicular is evidently l x l 2 + m x m 2 + n x n 2 = 0, and 
the condition for parallel lines is that l x = l 2 , m x = m 2 , and 
rii = n 2 . It can be shown by using vectors that the relation 
lj 2 + m x m 2 -f- n x n 2 = 1, combined with If + mf -4- nf = 1 and 
h 2 H- m 2 2 4- w 2 2 = 1 reduces to l x — l 2 , m 1 = ra 2 , and 7i 1 = n 2 . 

9. First-degree equation. — We now show that the equation 

(a) Ax + By+Cz + D^0 

represents a plane. A plane is, by definition, a surface which 
is such that the straight line joining any two points in the 
surface lies wholly on the surface, i.e. any other point on the 
line is also on the surface. 

Let Pi(x if ?/], Zx) and P 2 (x 2 , y 2 , z 2 ) be any. two points which 
satisfy equation (a). 

(b) .-.Ax^Byy+Czi+D^O. 

(c) Ax 2 + By 2 +Cz 2 + D = 0. 

Let P 3 (x 3 , y 3 , 2 3 ) be any other point on the line joining 
P(x ly yi, Zy) to P 2 (x 2 , y-2> %) an d let this point divide P,P 2 into 

segments such that x 3 = — • 
P 3 P 2 A* 2 

Then, x 3 , y 3 , and z 3 may be written 

ry. fc 2 Xi -p t i\X 2 k 2 y, + ~k\V2 „ _ k &\ + k\Z 2 

- 3 — — 7, I r. — ' 2/3 — , , — > z 3 — , , — * 

rC\ ~j- rC 2 K-, -j— f€ 2 K\ ~\~ >^2 

Substituting these values in the left-hand member of equa- 
tion (a), we have 

"'l + #2 #1 ~T" &2 "-1 ~T" #2 

which may be written, by rearrangement of terms, 

-A- (4^ + By, + C5* + Z>) + -A- (4^ 2 + By 2 + CM" D) I 

Ki -r fC 2 "'l ~r "^2 



SOLID GEOMETRY: POINTS AND LINES 463 

but this expression, by (b) and (c) above, is zero ; hence the 
point P 3 (x z , y 3 , z 3 ), any point on the straight line joining P x to 
P 2 , which points are on the locus of (a), is also on this locus. 

.-. Ax -f- By -4- Cz + D = represents a plane. 

The converse proposition is demonstrated in section 5 of the 
next chapter. 

A plane to pass through three given points is determined by 
substitution of the three points in equation (a) and solving for 
three of the constants, e.g. A, B, and C, in terms of the 
fourth. If the fourth constant chosen happens to be zero, 
another selection must be made. 

Illustrative problem. — Find the plane through (4, 4, 4), 
(3, 0, -5), and (0, -3,-7). 

Let Ax + By 4- Cz + D — represents the equation of the plane. 
Substituting, 

x 3 (1) 4J. + 4B + 4C + D = 0. 

(2) 3 A _.5O+D = 0. 

x 4 (3) - 3 B - 7 C + D = 0. 

Since it happens that only A, C, and B occur in (2), eliminate B 
between (1) and (3) by multiplying (1) by 3 and (3) by 4 and adding, 
obtaining 

(4) 12 A - 1(3 C + 7 B = 0. 

-4 (2) 3 A- 5 C+ B = 0. 

Eliminate A between (4) and (2) by multiplying (3) by — 4 and add- 
ing to (4) . This gives 

(5) 4 C + 3 B = ; C = - f B. 

Substituting in (2) the value of C found gives 

3 A + V- B + B = 0, A = - if B. 
Substituting value of C in (3) gives 

-3 B + V D + D = ; B = \% B. 
The equation of the plane may be written, 

- if Dx+ || By - | Bz + B = 0, or 
- 19x + 25?/-9z + 12 = 0. ^Ins. 



464 UNIFIED MATHEMATICS 

PROBLEMS 

1. Find the angle between the two lines of problem 5 of 
the preceding list of problems. 

2. Find the direction cosines proportional to 3, —2, and —5 ; 
find those proportional to 2, 3, and — 4 ; find the angle between 
two lines having these direction cosines that you have found. 

3. Find the equation of a plane whose intercepts on the 
axes are, respectively, 3, — 5, and 7. 

4. Find the equation of a plane through the points (3, 0, 5), 
(-2, 11, 7), and (0, 11, 7). 

5. The parametric equations of any line through (3, 0, 5) 
can be written 

x — 3 = r cos a, y — = r cos /?, z — 5 = r cos y. 
If this line is to pass through (—2, 11, 7), these coordinates 
must satisfy these equations. Make the substitutions, re- 
spectively; square and add the corresponding numbers and 
thus obtain the value of r, giving the distance of the point 
(3, 0, 5) from (—2, 11, 7). Find then the values of cos a, 
cos (3, and cos y. 

6. If a rectangular box with sides parallel to the coordinate 
planes has the line joining (3, 0, 5) to (— 2, 11, 7) as a principal 
diagonal, find the lengths of the sides, the length of the 
diagonal, and so find the direction cosines of the line joining 
the two points. 

7. Discuss the loci of the following equations and find three 
points on each locus : 

a. 3 a + 11 = 0. 

b. a 2 - 9 = 0. 

c. z-5 = 0. 

d. x — y — 5 = 0. 

e. z-2 y + 10 = 0. 

/. x + 2y + 3z-8 = 0. 

8. Where do points lie which are common to the loci of the 
two following equations : x = 3, and y=-5? 

<c-3 = 0andz-22/ + 10 = 0? 



CHAPTER XXXI 

SOLID ANALYTICS; FIRST-DEGREE EQUATIONS AND 
EQUATIONS IN TWO VARIABLES 

1. Locus of an equation in three variables. — Any equation 
involving three variables has for its locus a surface which 
may, in special forms of the equation, reduce to one or more 
lines or points. We obtain points on such a surface by giving 
values to two coordinates, e.g., x and y, and solving for the 
third, e.g., z. Thus we have found that any first-degree equa- 
tion represents a plane. 

Note, x 2 + y 2 + z 2 = represents only a point (0, 0, 0), or a point 
sphere. 

x 2 + y 2 = represents the z-axis since everywhere on this axis x = 
and y = 0. 

2. Intersections of loci. — (See Chapter V, Section 2.) Any 
point which satisfies two equations involving three variables 
lies, in general, upon a curve which is common to the two 
surfaces represented. 

When three equations are regarded as simultaneous, points 
of intersection of the three surfaces are obtained. Under 
special relations between the three given equations, these 
points may lie upon a line, but, in general, three simultaneous 
rational integral algebraic equations determine a finite num- 
ber of points of intersection. 

Just as a family of lines through the intersection of two 
given lines is obtained in Chapter V, Section 4, in the form 
^i + kh = 0, so the equation f x {x, y, z) + kf 2 (x, y, z) = 

465 



466 



UNIFIED MATHEMATICS 



represents a family of surfaces which pass through the inter- 
section curves of the two given surfaces. 

Thus, x 2 + y 2 + z 2 = 25 represents a sphere of radius 5 ; x 2 = 9 repre- 
sents two planes parallel to the 2/2-plane ; the equation 

x 2 + y 2 + z 2 ~ 25 - k(x 2 - 9) = 
represents for all values of. A; a surface through the intersections of the 
sphere and the plane. For k = — 1, this surface reduces to a cylinder, 

y 2 + z 2 - 16 = 0. 

3. Cylindrical surfaces. — Any equation in two variables, as 
x and y, represents in space a cylinder whose axis is parallel 
to the axis designated by the third variable. 




Cylindrical surfaces: 

Elliptic Hyperbolic Parabolic 

The curves indicated on these surfaces are cubic space curves. 

If an equation f(x, y) = is given in x and y, any point 
( x i> Vi) which satisfies the equation will lie upon the curve in 
the a:y-plane given by f(x, y)= 0. Considered as a point in 
space, the point (x h y x , 0) satisfies the equation, and further 
it is evident that (x x , y x , z), irrespective of the value of z, will 
also satisfy f(x, y) = 0, since the ^-coordinate does not enter 
at all. 



SOLID ANALYTICS 



467 



Thus, (o, 4, 0) satisfies the equation 

z 2 + y 2 - 25 = 
and also the points (3, 4, 1) or (3, '4, — 10) or (3, 4, 8,) will satisfy the 



equation 



x 2 + y 2 - 25 = 0. 



But all points (a^, y u z), for varying values of z only, lie on a 
parallel to the z-axis through (x ly y x ) and hence all points on 
the surface generated by a straight line moving parallel to the 
z-axis and touching the curve f(x } y) = 0, in the ay-plane, lie 
upon a cylinder. The curve f(x, y) = is called the directrix 
of the cylinder and the moving line is called the generator or 
element of the cylindrical surface. Similarly, when an equa- 
tion is given in x and z or in y and z, a cylinder is represented. 
A plane given by a first-degree equation in two variables, 
or one variable, is a special case of the preceding. 

4. Straight line as the intersection of two planes. — Just as the 



equation 



y - y\ 



- represents in the plane the straight 



y 2 - y l x 2 - x l 
line joining P x {xi, y ± ) to P 2 (x 2 , y 2 ), so the three equations, 
sb — xi _ y — yi _ z^-zi 



«2 - ®i 2/2 - yi H — *1 

represent in space the 
straight line joining 

A0*i, Vu Zi) 
to P 2 (x 2) y,. z 2 ). 



There are three equal- 
ities which are ob- 
tained by leaving out 
in turn each of the 
fractions, but there 
are only two inde- 
pendent equations, as 



■ : , 


- 












I 1 r 






.. 


. 












P 


± 




— — 
















''*: 


HMz 


:) 






































































































. 




























































/ 
















> \T 




• / 










































^ 


sSQ^ 


4— 








-^- — 








1 


P7^ 




/ 



















■f- — / 


•/ 







— — 




















1 Wv 
















// w 








































Jg. 1-j' Ll_l 






















tm 




































r ; " 


L- 






































(Xi,?/-, 




































/ 




















/ 




















/ 




















/ 
































-HA- 


( 




1 i 
























_ 


~ 
















77' 




' i , 










' ! 


' 


I- 




tf£r\ 


J.J.1J 



















Line joining Pi to P 2 in space 



468 UNIFIED MATHEMATICS 

the third equality would follow always from the first two 
which were given. 

These formulas can be obtained directly from the properties 
of similar triangles, or from the parametric equations of 
section 5 of the preceding chapter. The latter method brings 
out the important fact that the values x 2 — x ly y 2 — y if and 
z 2 — Zi are proportional to the direction cosines of the given 
line, and the values themselves of these cosines can be 
obtained, using the fact that the sum of the three squares is 
equal to unity. The derivation of the theorems mentioned is 
left as an exercise to the student. 

The parametric forms of the equations of a straight line may 
be written, 



x — x-^ 
cos a 


COS 


2/i. 


_z — 

COS 


*1_ 
y 


X — Xi _ 


y- 


■Vi 


z- 


-Zi 



or 

r 
k cos a k cos /3 k cos y k 

Further, any equations which can be put in one of the two 
forms above represent a straight line, and the denominators of 
the fractions are proportional to the direction cosines of the 
line. . 

The equations of the straight line in the form 

x — x x _ y — y x _ z — Zx 
a o c 

wherein a, b, and c are necessarily proportional to the direction 
cosines of the line, are called the standard or symmetrical 
equations of the line. 

In general, any curve in space is given as the intersection of 
two surfaces by the equations of the two surfaces. In particu- 
lar, the straight line is given by the equations of any two 
planes which pass through the line. Of the infinite number 
of planes, the pencil of planes, which pass through a given 
line; the three planes, called projecting planes of the line, 
which are parallel to the coordinate axes are of particular 



SOLID ANALYTICS 469 

importance. These equations will evidently be first-degree 
equations in two variables. In the standard form the equality 
of any two members gives one of the projecting planes through 
the given line. 

Illustrative problem. — Find the direction cosines of the 
straight line determined by the two planes 

(a) x + y>— 3z— 5 = 0. 

(b) 3x-y — 5z — ll = 0. 

Find the projecting planes parallel to the coordinate axes (or perpen- 
dicular to the coordinate planes) ; find the points where this line pierces 
the coordinate planes. 

Any plane through the line of intersection is given by 

(c) x + y - 3 z— 5 + &(3 x — y—5 z — 11)= 0. 

Giving to k the value — f, which is equivalent to multiplying (a) by 5, 
and (6) by — 3 and adding, and simplifying you have, 

5 x + 5 y — 15 z — 25 — 9 x + 3 y + 15 z + 33 = 0, or 
(c?) — 4 x + 8 ?/ + 8= 0, as the plane of projection on the xy-plane. 
Eliminating y, k — 1, gives, 

(e) 4 x ^- 8 z — 16 = or £ — 2 z — 4 = 0, the plane of projection on 
the xz-plane. 

Eliminating x, k =— £, gives, 

(/) — 4*/ + 4z + 4 = 0, which might have been obtained from (d) 
and (e), the plane of projection on the yz-plane. 
Solving for x, in (d) and (e), 

x = 2(?/ + 1) and 

x = 2(z +2). 

x = 2(y+l)=2(z + 2). 

fl — o _ y + l _ z + 2 

2 ~ 1 "" 1 
The denominators 2, 1, and 1 are proportional to the direction cosines of 
this line. Hence 

cos a = 2 m, cos £ = 1 m and cos 7 = 1 m, 

giving. 

cos 2 a + cos 2 /3 + cos 2 7 = 4 m 2 + ra 2 + m 2 = 1 ; 6 m 2 = 1 ; m = ± — . 

Either sign may be taken, but for convenience, make cos a positive. 

2 11 

cos « = — - , cos p = — - , cos 7 = — - , the direction cosines. 

V0 V6 v 6 



470 



UNIFIED MATHEMATICS 



To find where this line pierces any coordinate plane, as z = 0, solve 
the equation of the coordinate plane as simultaneous with the two given 
planes which determine the line. 

This gives here (4, 1,0) as the piercing point with the ajy-plane. 
Similarly we find the intersection with any plane. 

A parallel line to our given line through a given point would be de- 
termined by two planes through the given point parallel to the two given 
planes which determine the line. Why ? Determine the parallel line 
through (1, - 5, 6). 



5. Normal form of the equation of a plane. — (See Section 3, 
Chapter IX.) In the plane, the equation x cos a+y sin a— p=0, 
which may be written x cos a + y cos /? — p = 0, represents 
the equation of a straight line in normal form, which line is 
such that the perpendicular from the origin upon it has the 
length p and makes the angles a and /3 with #-axis and ?/-axis. 
Similarly, in space, the equation 

x cos a -f- y cos j3 + z cos y — p — 
represents a plane which is such that the perpendicular from 
the origin upon it has the length p and makes the angles a, 

fi, and y with the 
#-axis, y-Sixis, and 
z-axis respectively. 

Evidently, if a 
plane is given and 
a perpendicular ON 
of length p } having 
direction cosines a, 
/3, and y, is dropt 
from the origin to 
this plane, the point 
JVis (p cos a, p cos ft, 
p cos y) and the 
extension of the per- 
pendicular by the length p gives the point N f (2 p cos a, 
2p cos/?, 2p cosy). Any point P(x, y, z) which is equidis- 
tant from O(0, 0, 0) and N'(2p cos a, 2 p cos j3, 2 p cosy) 




ON of length p; ON' of length 2p 
Direction angles of ONN' : a, P, y 



SOLID ANALYTICS 471 

lies on our plane. Writing and. equating these distances, we 
have, 

a* 4. yi 4. 2 2 _££ _ 2p cos a) 2 +(y — 2p cos /?) 2 + (2 — 2p cos £) 2 . 

Whence, 

(4 p cos «)# + (4p cos (3)y + (4 p cos y)z 

= 4p 2 (cos 2 a + cos 2 ft + cos 2 y), 
giving finally 

jc cos a + 2/ cos /? -f- z cos y — p = 
as the equation. 

In the plane the distance from any point (x lf y x ) to a line is 
obtained by writing the equation of the line in normal form and 
substituting therein for x and y, x x and y ± . In space the dis- 
tance of a point (x ly y l} z x ) from a plane is obtained by writing the 
equation of the plane in normal form and substituting therein 
these coordinates of the point for x, y, and z, respectively. 

To reduce a linear equation to normal form, you divide the 
equation through, after transposing all terms to the left-hand 
member, by the square root of the sum of the squares of the 
coefficients of x, y, and z, choosing the sign opposite to the sign 
of the constant term. The proof is not similar to the proof 
of the corresponding theorem in plane analytics. 

Parallel planes are represented by linear equations having 
the corresponding coefficients, of x, y, and z, equal or propor- 
tional. 

PROBLEMS 

1. Put the following equations in normal form and deter- 
mine the distance of each plane from the origin : 

a. 2x-3y+4:Z-ll=0. 

b. x -j- y -f- z — 5 = 0. 

c. 2 x - 3 y - 11 = 0. d. z - 7 = 0. 

Determine the direction cosines and the direction angles of 
the normals to each of the above planes. 



472 UNIFIED MATHEMATICS 

2. Find the equations in standard form of a line from 
(1, 2, 5) perpendicular to the first plane in problem 1 ; 
through (0, 0, 0) perpendicular to the second plane in problem 
1; through (—2, —3,4) perpendicular to the third plane in 
problem 1. Determine in each of these three problems the 
intersection of the perpendicular with the plane. 

3. Find the piercing points with the coordinate planes of 
the following lines : 

a. 2 x — 3 ?/ + 4 z — 11 = and x — y -\-z — 5 = 0. 

b. 2 x — 3 y 4- 4 z - 11 = and z — 7 = 0. 

c. 2 x - 3 y - 11 = and z - 7 = 0. 

4. Put the three lines of problem 3 in standard form. 
Note that in the second and third cases, since the given line 
lies in a plane parallel to the x?/-plane, the line makes an angle 
of 90° with the z-axis, i.e. cos y = 0. The equations of the sec- 
ond and third lines in standard form would have a zero de- 
nominator, and so it is better to put these equations in the form 
given in the third of these problems. The values of cos a and 
cos /3 are determined here from the equation 2 x — Sy — 11 = 0, 

2 3 
giving cos a = and cos B = ■ 

VIS VI3 

5. Find the equations of the straight lines through the 
two points : a . ( 3> 5} _ 2 ) and (0, 0, 7). 

b. (3, 5, - 2) and (0, 0, 0). 

c. (3,5, -2) and (-3,5, +2). 

6. Find the angle between the lines 

x + 4 _ y _ z — 3 _, x — 2_y_ z — 1 
2 ~ ^1 ~ - 3 -2 ~5~ 7 

7. Do the two lines in problem 6 intersect ? How can you 
determine whether any two given lines intersect ? Note that 
the problem is entirely analogous to the problem in plane 
analytics as to whether three given lines intersect, and is 
solved in the same manner. Write the equations of two 
intersecting lines. 



SOLID ANALYTICS 



473 



8. Determine the curve in which the sphere 
X 2 + y i + Z 2 _ 400 = 

is intersected by the plane y — 9 = 0. Xote that substituting 
y = 9 is equivalent to writing 

a-2 + y2 + ^ _ 25 -(?/ + 9)(y - 9) = 0, 



■■ 





: : 


■ 


1 ■ 


i 




Mil 


















































































■^ 






^N 










y 


















X 


















yf 


















/ 
















=7= 








w 










4«c 












^rii 






/^ 


















/ 


^^^ 
















/ 








/ 




SJ 






sS 








f 




^\ 






s^ 
























£. 






V 










































O 






































, . i 


















/ 


















1 ! , / 


















1 i ! 1 / 


















1 / 


















1 / 


















/ 


















/ 


















/ 


















/ 


















/ 


















/ 








X, 










s , 








\_ 










X 




























~*-<^ 










' 







































Sphere cut by a plane 

which gives, of course, a new surface passing through the 
intersection curve of the first two surfaces. 

9. Find the intersection of the sphere x 2 + y 2 + z 2 — 100 = 
and the cylinder x 2 + y 2 — 36 = 0. 

10. L'pon what cylinder, parallel to one of the coordinate 
axes, does the intersection of the plane x = 5 with the surface 
x 2 + ±y 2 = 2oz lie? 

11. Find the intersection of the line 

x - 1 = 2 r, y-2 = 3r, z + 3 = - 5 r 

with the sphere x 2 + y 2 + z 2 — 100 = by substituting these 
values in the equation of the sphere and solving for r. Xote 
that since the right-hand coefficients, 2, 3, and — 5, are not the 



474 UNIFIED MATHEMATICS 

direction cosines of this line, but only proportional to them, 
the values of r obtained are not the distances from (1, 2, — 3) 
to the points of intersection with the sphere, but are propor- 
tional to these distances. The points of intersection are 
obtained by substituting the values of r found back in the 
equations of the line and solving for (x, y, z). 



Hyperboloid of one sheet Hyperboloid of two sheets 

CHAPTER XXXII 

SOLID ANALYTICS: QUADRIC SURFACES 

1. General equation. — In plane coordinates, any equation of 
the second degree represents a conic section. Similarly, in 
space coordinates, any equation of the second degree repre- 
sents a quadric surface. The types of quadric surfaces, 
limited in number, are closely allied to the types of conic sec- 
tions. In plane analytics, it is shown that the general equation 
of the second degree, containing the " cross-term " xy, intro- 
duces no new curves, only the same curves, represented by 
the different types of equations in which no xy-teim appears. 
It is likewise true in space that the general equation contain- 
ing any or all of the " cross-terms," yz, xz, and xy, presents 
no surfaces different from those which may be represented 
by the general equation containing no cross-term. 

Methods of transformation of coordinates quite similar to 
those discussed in Chapter XXIV apply to space coordinates, 
but the limitations of a first course preclude any discussion of 
the methods and results. 

Any surface given by an equation of the second degree is 

475 



476 UNIFIED MATHEMATICS 

cut by any plane in some form (including, of course, limiting 
forms) of conic section. The coordinate planes very evidently 
cut any quadric surface in a conic, since the curve of inter- 
section in the coordinate plane is given by an equation of the 
second degree in the two variables of that plane. The trans- 
formations mentioned above are desirable for the general proof, 
but another method is indicated below. 




Ellipsoids : 
q Prolate Oblate General 

Spheroid Spheroid ' Ellipsoid 

2. Ellipsoids. — The equation of a sphere has been given as 
An ellipsoid is given by the equation 

(x-hy (y-Jcy (z-iy = 1 

a 2 6 2 c 2 

This surface is related to the three spheres, 

(x - hy + ( y - ky+(z - iy = a 2 , ' 

(i_ft)I + ( y _fc)l + (g_i)l S=s 6», 

(a- -hy +( y -ky +(z- iy = c 2 , 

very much as the ellipse is related to its auxiliary circles. 

The parametric equations of the above ellipsoid are 

x — h = a cos a, 
y — k = b cos /?, 
z — I = c cos y. 



SOLID ANALYTICS: QUADRIC SURFACES 477 

The elimination of a, (3, and y, employing 

cos 2 a + cos 2 /3 + cos 2 y = 1, 

gives the equation of the ellipsoid in the standard form above. 

The quantities a, b, and c represent the semi-axes of the 

ellipsoid. If two of these denominators are equal to each 



— 


=* 


r 


V- 

A 

\ 


^ 


t 


— Y> 
B 


— v— 

N 


\ 




\ ' i i 





i 


^ 






*H 








3^ ( 


=4=M-Z"- 


\ ' \ 




— -1- 


M 






x: 

7 




^ 


I) 


r 
l 


1 \| 


1 

.1/ , 


i i I I 


\ 




=^ 


— 


-/- 

/ 

/ 


-tX( 




/ 


7- — - 


k 


— — r 
i 


i 
























■fm- 




— — 


— - 












[— 






M — 



Ellipsoid of revolution, with x-axis as axis of revolution 

other, the ellipsoid is an ellipsoid of revolution about an axis 
parallel to the axis corresponding to the term with the odd 
denominator. 



Thus. 



25 16 16 



x 2 



is an ellipsoid of revolution, obtained by revolving the curve — + 



about the x-axls. 



25 16 



The derivation of the formula of the ellipsoid of revolution 
is as follows, PN* + NM 2 = PM 2 , but QM=PM, radii of 
the circle QPR about M, with lettering as indicated on diagram 
given above. Now for all points on this circle the ^-coordinate 
is the same, 

f + z 2 = PM 2 == QM* = 16^1 - — 
which is a relation true for every point on the circle obtained 



478 UNIFIED MATHEMATICS 

by rotating the point Q on the ellipse h *— = 1 about its 

25 16 
axis. But Q is any point on the ellipse, and hence P may be 
any point on the surface obtained by revolving the ellipse 
about its axis. 

Hence, for every point on this surface, 

2/2 _j_ Z 2 _ le/'l . 



or 



25 16 16" ' 



Any ellipsoid of revolution obtained by revolving an ellipse 
about its major axis is called & prolate spheroid, and is shaped 
like a football ; an oblate spheroid is obtained by rotating an 
ellipse about its minor axis, and is shaped like a circular 
cushion or the earth. 

PROBLEMS 

1. Find the equation of the sphere having the center at the 
origin and passing through the point (—2,5, 6). Give the 
seven points which lie on this sphere and are symmetrically 
situated to the given point with respect to the coordinate 
planes. 

2. Find the equation of the preceding sphere if the center 
is at (3, — 2, 12). Find by using conditions of symmetry with 
respect to planes through the center parallel to the coordinate 
planes seven further points on this sphere. 

3. Write the equation of the ellipsoid having the center at 
the origin and semi-axes equal to 2, 3, and 5 respectively (x, y, 
and z order). Find three points on this ellipsoid. Write the 
equations of three circle's which lie on this surface. Write the 
equations of the traces on the coordinate planes, i.e. the inter- 
sections with these planes. Draw the graph. 

4. If a football is 10 inches long with a diameter of 8 
inches, write the equation of the surface, assuming it to be an 
ellipsoid. Draw the graph to scale. 



SOLID ANALYTICS: QUADRIC SURFACES 479 

5. Assuming that an air cushion 18 inches in diameter and 
6 inches high is an ellipsoid, write the equation of the surface. 
Draw the graph to scale. 

6. Find the six principal foci of the ellipsoid in problem 3. 
These are the foci of the traces on the coordinate planes. 

3. Hyperboloids. — By rotating the hyperbola 
tf_y 2 =1 
a 2 b 2 
about either axis, a hyperboloid of revolution is obtained. 
Rotation about the principal axis, the #-axis here, gives a 
surface of two separated parts, called a hyperboloid of revolution 
of two sheets. The equation is, 

^ 2 _^2_^2_ 1 
a 2 b 2 b 2 ~ 
The method of derivation, which we outline, is general, and 
being applied to the surface obtained by revolving any curve, 
V =f( x ) about the #-axis, will give the equation of the surface 
in the form y 2 + z 2 = \_f(x)~\ 2 . 



Given sL = 1 revolved about the a?-axis. 

a 2 b 2 







X J^"* *i 


4- SjL?l 


+■ *' *+ t 


T " "'^/' \f 


x S tt i 


g:g::::::::::;^:::_^^:::::|::::::::::::::::::: 




II Nil ' / u 


iS\< ' ' / 1 ■ - ------ 


- — h 'sS 1 1 i ; h — r""T 5 


— ±0 - ,\\- |lii/: = L:^::::::::::::::: 


/ 'li III "V, \ 'v c' 


'/I ' ' ' n^"" 4 - "^ " ~~ ~ 


/ ■ i M 1 ill! \ £,_---—- 




;;:!;!!! T 


~N+4fN-4fT + -T^ 



Hyperboloid of two sheets, of revolution 

The curves are slightly distorted. 



480 



UNIFIED MATHEMATICS 



Any point P(x, y, z) on this surface is obtained by the rotation 
of a point Q(x, y, 0) abont the a?-axis. 

The point Q generates a circle in a plane parallel to the yz- 
plane, in which x has everywhere the value given by OM. 

The equation of this 
circle is 

y* + z* = r 2 = PM 2 ==QM 2 . 

This radius r is evidently 
a function of x, being 
defined by the original 
equation given ; hence, 

r 2 = (ordinate on the hy- 
perbola) 2 = b*(— 2 - 1\ 

Hence the point P(x, y, z) 
satisfies the equation 



2/2 + 2:2 = 52, 




or, by rearrangement of 
terms, 






i, 



Hyperboloid of two sheets 

?/-axis as principal axis. 



the hyperboloid of revolu- 
tion of tivo sheets. 

Note that precisely this 
surface would have been 
obtained by revolving 



— — = 1 in the a^-plane about the aj-axis. 






Similarly, the hyperboloid of revolution of one sheet is obtained 
by revolving a hyperbola about its conjugate axis. The pre- 
ceding hyperbola revolved about the y-axis gives 



SOLID ANALYTICS: QUADRIC SURFACES 481 






= 1. 



«- 



The student will note that the axis of rotation in each case 
is given by the odd term. 

Corresponding to these 
surfaces of revolution are 
the general hyperboloids, 



a 2 



hyperboloid of two sheets, 
and 



+-=i, 




Hyperboloid of revolution, one sheet 

?/-axis as principal axis. 



hyperboloid of one sheet, 
which represent in each 
case a surface having a 
principal axis parallel to 
the axis of the odd term, 
e.g. the first has the x-axis 
as principal axis and the 
second has the ?/-axis as 
principal axis. Chang- 
ing the principal axis to another of the coordinate axes inter- 
changes two of the algebraic signs in the equation. 

4. Paraboloids. — By revolving the parabola y 1 = 4 ax about 
its axis the surface if + z 1 = 4 ax is obtained. 

This is called a paraboloid of revolution, or a circular parabo- 
loid, and is the type of surface which is fundamental in 
theater and auditorium construction. The derivation of the 
equation is left as an exercise for the student. 

The elliptic paraboloid is given by the equation 

* 2 + ^ = 4z ; 
a 2 6 2 

and sections parallel to the #?/-plane are ellipses. 



482 



UNIFIED MATHEMATICS 




Elliptic paraboloid 



Hyperbolic paraboloid 




The corresponding 
standard forms with 
y-axis and with a>axis, 
respectively, as princi- 
pal axis are, 



+ 



4*/, 



and 



•— + - = 4 x. 
b 2 c 2 



Elliptic paraboloid of revolution 



The equation 

a 2 b 2 

gives the most compli- 
cated of the quadratic 
surfaces, the hyperbolic 
paraboloid, a saddle- 
shaped surface which 



SOLID ANALYTICS: QUADRIC SURFACES 483 

is here represented by a photograph of a model of such a 
surface. 

The hyperbolic paraboloid may be generated by the motion 
of a given parabola, x 2 — 4 a 2 z = 0, moving parallel to the xz~ 
plane and having its vertex moving on the parabola 

y 2 =— 4 b 2 z. 

5. Cones. — If any straight line is revolved about another 
straight line in its plane as an axis, a cone of revolution is 
generated. Limit- 
ing forms are the 
cylinder, when the 
revolving line is par- 
allel to the axis, a 
plane when the re- 
volving line is per- 
pendicular to the 
axis, and a straight 

line when the revolv- _ . «. .*.*,. 

Cone generated by a straight line rotating 

ing line coincides about an axis in the same plane 

with its axis. 

Let y — -x revolve about OX. The cone of revolution gen- 




erated has the equation 



t 



0. 



The cone is itself a limiting form of the hyperboloids, as 
will be noted below. 

The general equation of the cone, whose axis is the ?/-axis 
and whose vertex is the origin, is 



±: _ yl _i_ z i 

a 2 b 2 c 2 



= 0. 



6. Conic sections. — - The method which we will here outline 
to prove that every plane section of a cone may be given by 



484 



UNIFIED MATHEMATICS 




Conic sections 



Ellipse Parabola • Hyperbola 



an equation of the second degree applies to a cone having an 
ellipse, parabola, or hyperbola as a base as well as to the cir- 
cular base, which is taken for convenience. 



/V»2 np. 2~ 

Given the cone •- = 0. 

a 2 b 2 V- 



Evidently, any one of the planes given by x = k cuts this 




Cone cut in an ellipse by a plane 



cone in a circular section, with a point as limiting case when 
x = 0. 



SOLID ANALYTICS: QUADRIC SURFACES 485 

The planes y = k and z = k cut this cone in hyperbolas. 
When y = or 2 = 0, the hyperbola " degenerates " into two 
straight lines intersecting at the vertex. 

The plane y = mx + k cuts the cone in a curve, which we 
will refer in this plane to the line y = mx + k, intersection of 
the £?/-plane and the cutting plane, and the line y = k, the 
intersection of the ?/z-plane and the cutting plane, as axes 
(of and z') of coordinates. From any point P(x, y, z) on the 
curve of intersection, drop a perpendicular PM to the sea- 
plane. The intersection curve satisfies the equation 

x 2 _ (mx + k) 2 _ ^ _ a 
a 2 b 2 b 2 ~~ ' 

or b-x 2 - a 2 (mx + k) 2 — a 2 z 2 = 0. 

Evidently, PM =z = z f , since PM is drawn in one of two per- 
pendicular planes, and perpendicular to the other. 





' ' r- 




3j •?? 




Ti « 




■*S--fi 
































i -^ 




i v ^S - 


















M 


-;0) ' ' ' P\X ]/'! — f P £ \I 




~ ■ ; ! > 1 ' -V f-^-f f- 




- 1 .y 








Tl P./ \ 




! ! 1 1 i i Ml! iM ! ! I 1 j I 1 -M,T,7/,2) 




















■ My', i 




' 6 ' 1 hi 




1 ' -J 



The elliptic section depicted in its own plane 



Further, BM = x' — sc Vl + m 2 , whence x 



stituting, we have, 
b 2 x' 2 a' 



Vl + ra 2 



2 a%m 



1 + m 2 ! + W 2 vl + m2 



a 2&2 _ a 2 z /2 = 



Sub- 



(6 2 - a 2 m 2 )x' 2 - a 2 z' 2 (l + m 2 ) - 2 a 2 fcm Vl + m 2 x - a 2 k 2 (l + m 2 ) = 0. 



486 



UNIFIED MATHEMATICS 



But this is an equation of the second degree in x 1 and z\ Fur- 
ther, the coefficient of x' 2 is (b 2 — a 2 m 2 ) and of z' 2 is — a 2 (l+m 2 ); 

hence the curve is an ellipse if m 2 > — , a parabola if m 2 = — , 
b 2 <# a 2 

and a hyperbola if m 2 < — 
a 2 

When the cutting plane has the form y = mx + nz -f Jc, the 

proof is more complicated but not essentially different. 

Every section of a cone may be represented by an equation of 
the second degree in two variables. 




Ellipse, parabola, hyperbola, and two straight lines as intersections of a 
cone by a plane 

PROBLEMS 

1. Name and discuss the following surfaces : 

a. x 2 -{- y 2 + z 2 -100 =0. 

b. x 2 + 2?/ 2 + 3z 2 -100 =0. 
a x 2 + 2y 2 -±z 2 -100 = 0. 

d. x 2 + 2y 2 -100 =0. 

e. x 2 + 2y 2 — 4z 2 =0. 
f a 2 -22/ 2 -4z 2 -100 =0. 



g. x 2 

h. x 2 -2y 2 



-100 =0. 
-100^ = 0. 



SOLID ANALYTICS: QUADRIC SURFACES 487 

L x*-2y 2 = 0. 

j. x? + 2y* = 0. 

Jc. x°- + 2if + 4:Z 2 = 0. 

/. x 2 + 2y 2 + <lz 2 + 100 =0. 

2. How would the addition of a term, 10 x, affect the locus 
of each of the preceding twelve expressions ? Discuss the 
change in each locus produced by changing the sign of x 2 in 
each expression from -f- to — . 

3. Find the equation of the cone obtained by revolving the 
line in the ay-plane y = 4 x — 10 about the ic-axis ; about the 
?/-axis ; about the z-axis. 

4. Find the "equation of the paraboloid obtained by revolv- 
ing the parabola z 2 = 8 x about the o>axis ; find the equation of 
the surface obtained by revolving this surface about the 
z-axis. Why has the latter surface not received particular 
discussion ? 

5. Find the equation of the cone obtained by revolving the 
line y = 1 x — 10 about the line y = 6. Note that this differs 
from the problems which we have considered in the text only 
by a change of origin, or a transformation of the type 

x = x' + h t y=ij'-\-k, z = z' + l 

6. Find the locus of a point which is equidistant from the 
point (4, 0, 0) and from the plane £ + 4 = 0. What is the 
surface ? 

7. Find the locus of a point the sum of whose distances 
from the points (4, 0, 0) and ( — 4, 0, ) is constant and equal 
to 10. What is the surface ? 

8. Find the locus of a point the difference of whose distances 
from two points (4, 0, 0) and (—4, 0, 0) is constant and equal 
to 6. 

9. How would you find in space coordinates the distance 
from a point to a line ? Apply to finding the distance from 

(1, 3, - 5) to the line ^^ = £=J = z -^. 
J 3-14 



488 UNIFIED MATHEMATICS 

7. Limiting forms. — The limiting forms corresponding to 
the ellipsoid are given by equations of the type 

a* b* c 2 

o r («-^ + (y--fc)' + fe^jg = o 

a 2 b 2 c 2 

the first of which represents the point (0, 0, 0) and the 
second the point (Ji, k, I). 

The method of approaching this limit is best indicated by 
writing the equation as 

a 2 b 2 c 2 

As k approaches 0, the semi-axes Vka 2 , Vfc^, and -yfkc 2 ap- 
proach as a limit. 

In a similar way the hyperboloid equations approach, as 
limits, the equations representing cones asymptotic to the 
given hyperboloids. 

The limiting forms of the paraboloids are equations in two 
variables, and reduce to two planes, or to cylinders. 

The limiting forms of equations in two variables represent- 
ing cylinders correspond, with proper and more or less evident 
changes, to the limiting forms of the corresponding equations 
in plane analytics. 

Thus, any equation of the second degree f(x, y, z) = 0, 
whether in one or two or three variables, of which the left- 
hand member can be factored into two real linear factors in 
the variables, represents two planes which constitute also a 
type of quadric surface. 

8. Applications. — The applications of the conic sections 
which have been given in plane analytics are strictly applica- 
tions of surfaces, or solids having these surfaces as boundaries. 

Thus, a bridge having a parabolic arch uses a solid having a 
parabolic cylinder as bounding surface. 



SOLID ANALYTICS; QUADRIC SURFACES 489 

The equation of the paraboloid in the Hill Auditorium, with 
the foot as unit of length, is y 2 -f z 2 = 70.02 aj ; the skylight in 
the ceiling of the Hill Auditorium is bounded by an elliptical 
cylinder, 

— + ^— = 1, dimensions in feet. 
76 2 50 2 

"Any of the automobile reflectors are paraboloids, in general, of 
revolution. Hyperboloids are used as revolving cones in the 
manufacture of iron pipes ; these pipes are passed between 
two revolving cones whose axes are inclined at 90° to 
straighten the pipes. 

9. Circular sections. — Given the ellipsoid represented by 

a 2 b 2 c 2 

the question arises as to what planes cut this surface in 
circular sections. 

The method which we have given above, under section 7, 
for determining the nature of the curve cut out of the cone by 
the plane y = mx -f k applies to this problem. In the ellipsoid 
above planes y = k cut the surface in ellipses. The develop' 
ment as given shows that any plane y = mx cuts this surface 
in a curve given by the intersection, also, of the surface, 

x 2 . m 2 x 2 , z 2 = 1 



a 2 b 2 
and the given plane, or a curve, 



m^x 



2V2 



+ ,..„ ■ ...«x +^ = l> 



a 2 (l + m 2 ) b 2 {l + m 2 ) 

referred to the line j ^ and the line — ~ as axes of 
12 = i x = 

reference, both lying in the plane of the section. This curve 

can be written, 

c 2 (6 2 - a 2 m 2 )x' 2 + a 2 b 2 (l + m 2 )z' 2 = a 2 b 2 c 2 (l + m 2 ). 



490 UNIFIED MATHEMATICS 

Equating the coefficients of x' 2 and z' 2 gives 

9 = b 2 (a 2 -c 2 ) 
m a 2 (c 2 -b 2 y 



mss± bV*-* 

ay C 2 _ 52 






The two planes, y = ± a c x\ and all planes parallel to 

ttVc 2 -& 2 

them, cut this surface in circular sections. For these to be real 

planes c must be intermediate in value between a and b. If c 

is not intermediate between a and b, then planes either of the 

form y = mz or y = nx will make real circular sections. 

The method applies to elliptic cylinders, to elliptic cones, 
and to hyperboloids, as well as to the ellipsoid. 

A simpler method, assuming c as the intermediate value, is to 

x 2 v 2 
find in the ellipse — (- 2- = 1 a diameter of length 2 c ; this 
a 2 b 2 

diameter with the 2-axis determines the plane of a circular 

section. 

10. Tangent planes and tangent lines. — The formula 

Ax x x + By x y + G(x + x x ) + F(y + 3h) + C = 0, 

which gives the tangent to 

Ax 2 + By 2 + 2 Gx + 2Fy + C=0, 

at the point Pi(x x , y x ) on the curve applies in space analytics, 
with the addition of the corresponding z 2 and z terms, to 
give the tangent plane to the quadric surface at a point 
Pi(x x , y lf Zi) on the surface. 

The tangent plane at Pi(x lt y h z x ) to the surface 

Ax 2 + By 2 + Cz 2 + 2 Gx + 2 Fy + 2 Ez + K =0 

is given by the equation, 

Ax x x + By x y + Cz,z +' G(x + x,) + F(y + y x ) + E(z + z 1 ) + K= 0. 



SOLID ANALYTICS: QUADRIC SURFACES 491 

When the point Pi(x h y 1} z{) is not on the quadric surface, 
this equation represents not the tangent plane but the polar 
plane of the point (x u y u z x ) with respect to the surface. For 
any point outside of the surface, tangent planes to the surface 
have their points of tangency situated upon a plane, the polar 
plane of the point Pi(x l7 y ly z^. A more complete discussion of 
the polar plane would reveal many other points of similarity 
between the polar plane as related to its quadric surface and the 
polar line as related to its conic. 

The intersection of a tangent plane at a point Pi(x 1} y lf z x ) 
on the surface with any other plane through P ± gives a tan- 
gent line to the surface at P x . 

11. Ruled surfaces. Generating lines. — Any surface which 
can be generated by the motion of a straight line moving 
according to some law is termed a ruled surface. Evidently, 
by its method of generation, such a surface has straight-line 
elements, called rectilinear generators, which lie wholly upon 
the surface. 

Certain of the quadric surfaces are ruled surfaces. Evi- 
dently all the cylinders, the cones, and the pairs of planes 
belong in this class. The ellipsoid, being confined to a finite 
portion of space, does not have right-line elements lying 
wholly upon the surface ; nor do the elliptic paraboloid and 
the hyperboloid of two sheets have right-line elements. 

The hyperbolic paraboloid and the hyperboloid of one 
sheet do have rectilinear generators. We will find the equa- 
tion of the families of lines which lie wholly upon one of the 
surfaces in question ; the method will apply to the other ruled 
quadric surfaces. 

Any point upon the hyperboloid 

Bl _ t + t_ — 1 

a 2 b 2 c 2 ~~ 

very evidently satisfies the equation 

- - y - = 1 - - 
a 2 V c 2 ' 



492 



UNIFIED MATHEMATICS 



which may be written, 



a o \a o 



This indicates that any point which satisfies the pair of linear 
equations ,_ ?= , 

a b \ 



a b k 

will satisfy the equation of our surface since it will make the 
product represented by the left-hand member of our equation, 

18,17,16,15 14 13 12 11 10 9 8 7 6 5 4 3,2 
19,20 2122 23 24 25 26 27 28 29 30 31,32,1 




30,31,32 12 34 5 6 7 89 10 11,12,13.14 

29,28 27 26 25 24 23 22 21 20 19 18 17,16,15 

The right-line generators on this hyperboloid of revolution are formed by 
connecting corresponding points on two circular sections 

An elliptic section is also indicated. 



in the second form above, equal to the product of the factors, 
representing the right-hand member. But every point which 



SOLID ANALYTICS: QUADRIC SURFACES 493 

satisfies the pair of equations for any given value of k lies 
upon a straight line, the intersection of the two planes given 
by the linear equations. Hence, every point upon this line 
lies upon the given surface for any value of k. 

It can be shown that no two lines of this family of lines, 
i.e. no two lines given by two values of k, intersect. 

Another family of lines also lies upon this surface. The 
equations of this second family of lines, with the parameter 
k, are as follows : 

y=kfi+ z - 






Every member of this family of lines can be shown to inter- 
sect every member of the preceding family and no member 
of its own family. 

PROBLEMS 

1. Find the equations of the rectilinear generators of the 
following surfaces : 

a. x 2 — y 2 — z 2 = 0. 

b. x 2 + y 2 — z 2 == 16. 

c. x 2 -y 2 — ±z = 0. 

d. x 2 — y 2 =0. 

2. Find the circular sections of the following surfaces : 



a. 


x 2 y 2 
25 16 


7 2 
-- = 1. 

9 


b. 


x 2 y 2 
25 16 


z 2 
-- =0. 
9 


c. 


x 2 +±y 2 


= 9 z. 


d. 


x 2 -f 4 y 2 


= 9. 


e. 


x 2 y 2 
100 36 


z 2 

- — = 1. 

16 



494 UNIFIED MATHEMATICS 

3. Write the equations of the tangent planes to each of the 
surfaces in the preceding problem at the point (x 1} y lt z x ) in each 
case upon the given surface. 

4. In problem 1 a, above, take fcj = 1 and k 2 = 2 and show 
that these two lines of the same family of rectilinear generators 
do not intersect. Write the second family of rectilinear gen- 
erators of the same surface and show that, taking fe = 1 (any 
other value would do), this line does intersect a given line 
Qd = 1) of the first set. How could you make this proof 
general ? 



TABLES 

Page 

Constants with their logarithms „ . , 496 

Squares and cubes of integers, 1 to 100 497 

Four-place logarithms of numbers, 100 to 999 498 

Four-place logarithms of numbers, 1000 to 1999 500 

Four-place logarithms of sines and cosines, 10' intervals .... 502 

Four-place logarithms of tangents and cotangents, 10' intervals . .' 504 
Four-place logarithms of sines, 0° to 9°, and of cosines, 81° to 90°, by 

minutes 506 

Four-place logarithms of tangents, 0° to 9°, and of cotangents, 81° to 

90°, by minutes • 507 

Four-place natural sines and cosines, 10' intervals 508 

Four-place natural tangents and cotangents, 10' intervals . . . „ 510 

Radian measure of angles 512 

Minutes as decimals of one degree ; e x and e~ x tables 513 

The accumulation of 1 at the end of n years 514 

The present value of 1 due in n years 515 

The accumulation of an annuity of 1 per annum at the end of 

n years 516 

The present value of an annuity of 1 per annum for n years . . . 517 
The annual sinking fund which will accumulate to 1 , or, by addition 

of i, the annuity which 1 will purchase 518 



495 



496 



UNIFIED MATHEMATICS 



Constants with their logarithms. 



Base of natural logarithms .... 
Modulus of common logarithms . 
Circumference of a circle in degrees , 
Circumference of a circle in minutes 
Circumference of a circle in seconds 
Radian expressed in degrees . . . 
Radian expressed in minutes . . . 
Radian expressed in seconds . . .. 
Ratio of a circumference to diameter 
7T = 3.14159265358979323846264338328 

Volumes and "Weights 

Cubic inches in 1 gallon (U S.) 
Gallons in 1 cubic foot ... 
Cubic inches in 1 bushel . . . 
Pounds per cubic foot of water (4 
Pounds per cubic foot of air (0° C 
Cubic feet in 1 cubic meter 
Cubic meters in 1 cubic yard 
Cubic inches in 1 liter . . 
Liters in 1 gallon (U. S.) 
Pounds in 1 kilogram . . . 
Metric ton in pounds . . . 
Volume of sphere, i -n-r 3 . . 



Ntjmbee 


Logarithm 


e = 2.71828183 


0.4342945 


u = 0.43429448 


9.6377843-10 


= 360 


2.5563025 


= 21600 


4.3344538 


= 1296000 


6.1126050 


= 57.29578 


1.7581226 


= 3437.7468 


3.5362739 


= 206264.806 


5.3144251 


tt = 3.14159265 


0.4971499 



Lengths and Areas 

Inches in 1 meter (by Act of Congress) = 39.37 
Feet in 1 rod, 16.5 ; yards in 1 rod . 

Square feet in 1 acre = 43560 

160 square rods = 1 acre ; 640 acres = 1 square mile ; 
3.281 feet = 1.094 yards = 1 meter. 



= 231 


2.3636 


= 7.48 


.8739 


= 2150.4 


3.3325 


= 62.43 


1.7954 


= 0.0807 


8.9069-10 


= 35.32 


1.5480 


= 0.76 


9.8808-10 


= 61.03 - 


1.7855 


= 3.786 


.5783 


= 2.2 or 2.205 


.3434 


= 2205 


3.3434 


= 4.1888 r 3 




4.1888 


.6221 


= 39.37 


1.5952 


= 5.5 




= 43560 


4.6391 



TABLES 



497 



Squares and cubes of integers, 1 to 100. 



Square roots and cube roots of 1 to 100. 



Reciprocals of 1 to 100. 





a 




a 








H 




H 








5 


3 


< & 


H H 






<• 


a 


3h 


H ^ 


I J 




J3 


pq 


P o 


« O 


o 5 




p 


« 


P ° 


a o 


o £ 




Ct 


P 


a ° 


P o 


H O 




C 


P 


a ° 


P o 


w O 




QQ 


Q 


oqPS 


Otf 


P?« 




w 


O 


mti 


Otf 


tf § 


n 


n2 


?l 3 


Vn 


3 i— 


1 


n 


?l2 


)l3 


\ f n 


3.'— 

V n 


lj 

.0196 


l 


1 


1 


1.000 


1.000 


1.0000 


51 


2,601 


132,651 


7.141 


3.708 


9 


4 


8 


1.414 


1.260 


.5000 


52 


2,704 


140,608 


7.211 


3.733 


.0192 


3 


9 


27 


1.732 


1.442 


.3333 


53 


2,809 


148,877 


7.280 


3.756 


.0189 


4 


16 


64 


2.000 


1.587 


.2500 


54 


2,916 


157,464 


7.348 


3.780 


.0185 


5 


25 


125 


2.236 


1.710 


.2000 


55 


3,025 


166,375 


7.416 


3.803 


.0182 


6 


36 


216 


2.449 


1.817 


.1667 


56 


3,136 


175,616 


7.483 


3.826 


.0179 


7 


49 


343 


2.646 


1.913 


.1429 


57 


3,249 


185,193 


7.550 


3.849 


.0175 


3 


64 


512 


2.828 


2.000 


.1250 


58 


3,364 


195,112 


7.616 


3.871 


.0172 


9 


81 


729 


3.000 


2.080 


.1111 


59 


3,481 


205,379 


7.681 


3.893 


.0170 


10 


100 


1,000 


3.162 


2.154 


.1000 


60 


3,600 


216,000 


7.746 


3.915 


.0167 


11 


121 


1,331 


3.317 


2.224 


.0909 


61 


3,721 


226,981 


7.810 


3.936 


.0164 


12 


144 


1,728 


3.464 


2.289 


.0833 


62 


3,844 


238,328 


7.874 


3.958 


.0161 


13 


169 


2,197 


3.606 


2.351 


.0769 


63 


3,969 


250,047 


7.937 


3.979 


.0159 


14 


196 


2,744 


3.742 


2.410 


.0714 


64 


4,096 


262,144 


8.000 


4.000 


.0156 


1.3 


225 


3,375 


3.873 


2.466 


.0667 


65 


4,225 


274,625 


8.062 


4.021 


.0154 


16 


256 


4,096 


4.000 


2.520 


.0625 


66 


4,356 


287,496 


8.124 


4.041 


.0152 


17 


289 


4,913 


4.123 


2.571 


.0588 


67 


4,489 


300,763 


8.185 


4.062 


.0149 


IS 


324 


5,832 


4.243 


2.621 


.0556 


68 


4,624 


314,432 


8.246 


4.082 


.0147 


19 


361 


6.859 


4,359 


2.668 


.0526 


69 


4,761 


328,509 


8.307 


4.102 


.0145 


20 


400 


8,000 


4.472 


2.714 


.0500 


70 


4,900 


343,000 


8.367 


4.121 


.0143 


21 


441 


9,261 


4.583 


2.759 


.0476 


71 


5,041 


357,911 


8.426 


4.141 


.0141 


82 


484 


10,648 


4,690 


2.802 


.0455 


72 


5,184 


373,248 


8.485 


4.160 


.0139 


23 


529 


12,167 


4.796 


2.844 


.0436 


73 


5,329 


389,017 


8.544 


4.179 


.0137 


24 


576 


13,824 


4.899 


2.884 


.0417 


74 


5,476 


405,224 


8.602 


4.198 


.0135 


25 


625 


15,625 


5.000 


2.924 


.0400 


75 


5,625 


421,875 


8.660 


4.217 


.0133 


26 


676 


17,576 


5.099 


2.962 


.0385 


76 


5,776 


438,976 


8.718 


4.236 


.0132 


27 


729 


19,683 


5.196 


3.000 


.0370 


77 


5,929 


456,533 


8.775 


4.254 


.0130 


28 


784 


21,952 


5.292 


3.037 


.0357 


78 


6,084 


474,552 


8.832 


4.273 


.0128 


29 


841 


24,389 


5.385 


3.072 


.0345 


79 


6,241 


493.039 


8.888 


4.291 


.0127 


30 


900 


27,000 


5.477 


3.107 


.0333 


80 


6,400 


512,000 


8.944 


4.309 


.0125 


31 


961 


29,791 


5.568 


3.141 


.0323 


81 


6,561 


531,441 


9.000 


4.327 


.0123 


32 


1,024 


32,768 


5.657 


3.175 


.0313 


82 


6,724 


551,368 


9.055 


4.344 


.0122 


33 


1,089 


35,937 


5.745 


3.208 


.0303 


83 


6,889 


571,787 


9.110 


4.362 


.0120 


34 


1,156 


39,304 


5.831 


3.240 


.0294 


84 


7,056 


592,704 


9.165 


4.380 


.0119 


35 


1,225 


42,875 


5.916 


3.271 


.0286 


85 


7,225 


614,125 


9.220 


4.397 


.0118 


36 


1,296 


46,656 


6.000 


3.302 


.0278 


86 


7,396 


636,056 


9.274 


4.414 


.0116 


37 


1,369 


50,653 


6.083 


3.332 


.0270 


87 


7,569 


658,503 


9.327 


4.431 


.0115 


3S 


1,444 


54,872 


6.164 


3.362 


.0263 


88 


7,744 


681,472 


9.381 


4.448 


.0114 


39 


1,521 


59,319 


6.245 


3.391 


.0256 


89 


7,921 


704,969 


9.434 


4.465 


.0112 


40 


1,600 


64,000 


6.325 


3.420 


.0250 


90 


8,100 


729,000 


9.487 


4.481 


.0111 


41 


1,681 


68,921 


6.403 


3.448 


.0244 


91 


8,281 


753,571 


9.539 


4.498 


.0110 


42 


1,764 


74,088 


6.481 


3.476 


.0238 


92 


8,464 


778,688 


9.592 


4.514 


.0109 


43 


1,849 


79,507 


6.557 


3.503 


.0233 


93 


8,649 


804,357 


9.644 


4.531 


.0108 


44 


1,936 


85,184 


6.633 


3.530 


.0227 


94 


8,836 


830,584 


9.695 


4.547 


.0106 


45 


2,025 


91,125 


6.708 


3.557 


.0222 


95 


9,025 


857,375 


9.747 


4.563 


.0105 


46 


2,116 


97,336 


6.782 


3.583 


.0217 


96 


9,216 


884,736 


9.798 


4.579 


.0104 


47 


2,209 


103,823 


6.856 


3.609 


.0213 


97 


9,409 


912,673 


9.849 


4.595 


.0103 


4S 


2,304 


110,592 


6.928 


3.634 


.0208 


98 


9,604 


941,192 


9.899 


4.610 


.0102 


49 


2,401 


117,649 


7.000 


3.659 


.0204 


99 


9,801 


970,299 


9.950 


4.626 


.0101 


50 


2,500 

™2 


125,000 


7.071 

,A7 


3.684 

3 AT 


.0200 

1 


100 


10,000 


1,000,000 


10.000 


4.642 


.0100 

1 



498 



UNIFIED MATHEMATICS 



Logarithms of numbers from 100 to 549. 



5 



10 


0000 


0043 


0086 


0128 


0170 


0212 


11 


0414 


0453 


0492 


0531 


0569 


0607 


12 


0792 


0828 


0864 


0899 


0934 


0969 


13 


1139 


1173 


1206 


1239 


1271 


1303 


14 


1461 


1492 


1523 


1553 


1584 


1614 


15 


1761 


1790 


1818 


1847 


1875 


1903 


16 


2041 


2068 


2095 


2122 


2148 


2175 


17 


2304 


2330 


2355 


2380 


2405 


2430 


18 


2553 


2577 


2601 


2625 


2648 


2672 


19 


2788 


2810 


2833 


2856 


2878 


2900 


20 


3010 


3032 


3054 


3075 


3096 


3118 


21 


3222 


3243 


3263 


3284 


3304 


3324 


22 


3424 


3444 


3464 


3483 


3502 


3522 


23 


3617 


3636 


3655 


3674 


3692 


3711 


24 


3802 


3820 


3838 


3856 


3874 3892 


25 


3979 


3997 


4014 


4031 


4048 


4065 


26 


4150 4166 


4183 


4200 


4216 


4232 


27 


4314 


4330 


4346 


4362 


4378 


4393 


28 


4472 


4487 


4502 


4518 


4533 


4548 


29 


4624 


4639 


4654 


4669 


4683 


4698 


30 


4771 


4786 


4800 


4814 


4829 


4843 


31 


4914 


4928 


4942 


4955 


4969 


4983 


32 


5051 


5065 


5079 


5092 


5105 


5119 


33 


5185 


5198 


5211 


5224 


5237 


5250 


34 


5315 


5328 


5340 


5353 


5366 


5378 







1 


2 


3 


4 


5 


35 


5441 


5453 


5465 


5478 


5490 


5502 


36 


5563 


5575 


5587 


5599 


5611 


5623 


37 


5682 


5694 


5705 


5717 


5729 


5740 


38 


5798 


5809 


5821 


5832 


5843 


5855 


39 


5911 


5922 


5933 


5944 


5955 


5966 


40 


6021 


6031 


6042 


6053 


6064 


6075 


41 


6128 


6138 


6149 


6160 


6170 


6180 


42 


6232 


6243 


6253 


6263 


6274 


6284 


43 


6335 


6345 


6355 


6365 


6375 


6385 


44 


6435 


6444 


6454 


6464 


6474 


6484 


45 


6532 


6542 


6551 


6561 


6571 


6580 


46 


6628 


6637 


6646 


6656 


6665 


6675 


47 


6721 


6730 


6739 6749 


6758 


6767 


48 


6812 


6821 


6830 


6839 


6848 


6857 


49 


6902 


6911 


6920 


6928 


6937 


6946 


50 


6990 


6998 


7007 


7016 


7024 


7033 


51 


7076 


7084 


7093 


7101 


7110 


7118 


52 


7160 


7168 


7177 


7185 


7193 


7202 


53 


7243 


7251 


7259 


7267 


7275 


7284 


54 


7324 


7332 


7340 


7348 


7356 


7364 



0253 0294 0334 0374 
0645 0682 0719 0755 
1004 1038 1072 1106 
1335 1367 1399 1430 
1644 1673 1703 1732 

1931 1959 1987 2014 
2201 2227 2253 2279 
2455 2480 2504 2529 
2695 2718 2742 2765 
2923 2945 2967 2989 

3139 3160 3181 3201 
3345 3365 3385 3404 
3541 3560 3579 3598 
3729 3747 3766 3784 
3909 3927 3945 3962 

4082 4099 4116 4133 
4249 4265 4281 4298 
4409 4425 4440 4456 
4564 4579 4594 4609 
4713 4728 4742 4757 

4857 4871 4886 4900 
4997 5011 5024 5038 
5132 5145 5159 5172 
5263 5276 5289 5302 
5391 5403 5416 $428 

6 7 8 9 

5514 5527 5539 5551 
5635 5647 5658 5670 
5752 5763 5775 5786 
5866 5877 5888 5899 
5977 5988 5999 6010 

6085 6096 6107 6117 
6191 6201 6212 6222 
6294 6304 6314 6325 
6395 6405 6415 6425 
6493 6503 6513 6522 

6590 6599 6609 6618 
6684 6693 6702 6712 
6776 6785 6794 6803 
6866 6875 6884 6893 
6955 6964 6972 6981 

7042 7050 7059 7067 
7126 7135 7143 7152 
7210 7218 7226 7235 
7292 7300 7308 7316 
7372 7380 7388 7396 

6 7 8 9 



To avoid inter- 


polation in 


the 


first ten 


ines. 


use tables on the 


following pages. 




22 




1 


2.2 




2 


4.4 




3 


6.6 




4 


8.8 




5 


11.0 




6 


13.2 




7 


15.4 




8 


17.6 




9 


19.8 




21 


20 


19 


1 2.1 


2.0 


1.9 


2 4.2 


4.0 


3.8 


3 6.3 


6.0 


5.7 


4 8.4 


8.0 


7.6 


5 10.5 


10.0 


9.5 


6 12.6 


12.0 


11.4 


7 14.7 


14.0 


13.3 


8 16.8 


16.0 


15.2 


9 18.9 


18.0 


17.1 


18 


17 


16 


1 1.8 


1.7 


1.6 


2 3.6 


3.4 


3.2 


3 5.4 


5.1 


4.8 


4 7.2 


6.8 


6.4 


5 9.0 


8.5 


8.0 


6 10.8 


10.2 


9.6 


7 12.6 


11.9 


11.2 


8 14.4 


13.6 


12.8 


9 16.2 


15.3 


14.4 


15 


14 


13 


1 1.5 


1.4 


1.3 


2 3.0 


2.8 


2.6 


3 4.5 


4.2 


3.9 


4 6.0 


5.6 


5.2 


5 7.5 


7.0 


6.5 


6 9.0 


8.4 


7.8 


7 10.5 


9.8 


9.1 


8 12.0 


11.2 


10.4 


9 13.5 


12.6 


11.7 


11 




12 


1 1.1 


1 


1.2 


2 2.2 


2 


2.4 


3 3.3 


3 


3.6 


4 4.4 


4 


4.8 


5 5.5 


5 


6.0 


6 6.6 


6 


7.2 


7 7.7 


7 


8.4 


8 8.8 


8 


9.6 


9 9.9 


9 


10.8 


j£ 


a 
o 




1 


"3 




a 


o 




V 






B 


.& 






"3 

s 




1 


3 




0) 


M d, 




"3 


•S3 





TABLES 



499 



Logarithms of numbers from 550 to 999. 



8 



55 


7404 


7412 


7419 


7427 


7435 


7443 


56 


7482 


7490 


7497 


7505 


7513 


7520 


57 


7559 


7566 


7574 


7582 


7589 


7597 


58 


7634 


7642 


7649 


7657 


7664 


7672 


59 


7709 


7716 


7723 


7731 


7738 


7745 


60 


7782 


7789 


7796 


7803 


7810 


7818 


61 


7853 


7860 


7868 


7875 


7882 


7889 


6? 


7924 


7931 


7938 


7945 


7952 


7959 


63 


7993 


8000 


8007 


8014 


8021 


8028 


64 


8062 


8069 


8075 


8082 


8089 


8096 


65 


8129 


8136 


8142 


8149 


8156 


8162 


66 


8195 


8202 


8209 


8215 


8222 


8228 


67 


8261 


8267 


8274 


8280 


8287 


8293 


68 


8325 


8331 


8338 


8344 


8351 


8357 


69 


8388 


8395 


8401 


8407 


8414 


8420 


70 


8451 


8457 


8463 


8470 


8476 


8482 


71 


8513 


8519 


8525 


8531 


8537 


8543 


73 


8573 


8579 


8585 


8591 


8597 


8603 


73 


8633 


8639 


8645 


8651 


8657 


8663 


74 


8692 


8698 


8704 


8710 


8716 


8722 


75 


8751 


8756 


8762 


8768 


8774 


8779 


76 


8808 


8814 


8820 


8825 


8831 


8837 


77 


8865 


8871 


8876 


8882 


8887 


8893 


78 


8921 


8927 


8932 


8938 


8943 


8949 


79 


8976 8982 


8987 


8993 


8998 


9004 







1 


2 


3 


4 


5 


80 


9031 


9036 9042 


9047 


9053 


9058 


81 


9085 


9090 


9096 


9101 


9106 


9112 


83 


9138 


9143 


9149 


9154 


9159 


9165 


83 


9191 


9196 


9201 


9206 


9212 


9217 


84 


9243 


9248 


9253 


9258 


9263 


9269 


85 


9294 


9299 


9304 


9309 


9315 


9320 


86 


9345 


9350 


9355 


9360 


9365 


9370 


87 


9395 


9400 


9405 


9410 


9415 


9420 


88 


9445 


9450 


9455 


9460 


9465 


9469 


89 


9494 


9499 


9504 


9509 


9513 


9518 


90 


9542 


9547 


9552 


9557 


9562 


9566 


91 


9590 


9595 


9600 


9605 


9609 


9614 


92 


9638 


9643 


9647 


9652 


9657 


9661 


93 


9685 


9689 


9694 


9699 


9703 


9708 


94 


9731 


9736 


9741 


9745 


9750 


9754 


95 


9777 


9782 


9786 


9791 


9795 


9800 


96 


9823 


9827 


9832 


9836 


9841 


9845 


97 


9868 


9872 


9877 


9881 


9886 


9890 


98 


9912 


9917 


9921 


9926 


9930 


9934 


99 


9956 


9961 


9965 


9969 


9974 


9978 



7451 7459 7466 7474 
7528 7536 7543 7551 
7604 7612 7619 7627 
7679 7686 7694 7701 
7752 7760 7767 7774 

7825 7832 7839 7846 
7896 7903 7910 7917 
7966 7973 7980 7987 
8035 8041 8048 8055 
8102 8109 8116 8122 

8169 8176 8182 8189 
8235 8241 8248 8254 
8299 8306 8312 8319 
8363 8370 8376 8382 
8426 8432 8439 8445 

8488 8494 8500 8506 
8549 8555 8561 8567 
8609 8615 8621 8627 
8669 8675 8681 8686 
8727 8733 8739 8745 

8785 8791 8797 8802 
8842 8848 8854 8859 
8899 8904 8910 8915 
8954 8960 8965 8971 
9009 9015 9020 9025 



6 



8 



9063 9069 9074 9079 
9117 9122 9128 9133 
9170 9175 9180 9186 
9222 9227 9232 9238 
9274 9279 9284 9289 

9325 9330 9335 9340 

9375 9380 9385 9390 

9425 9430 9435 9440 

9474 9479 9484 9489 

9523 9528 9533 9538 

9571 9576 9581 9586 
9619 9624 9628 9633 
9666 9671 9675 9680 
9713 9717 9722 9727 
9759 9763 9768 9773 

9805 9809 9814 9818 
9850 9854 9859 9863 
9894 9899 9903 9908 
9939 9943 9948 9952 
9983 9987 9991 9996 



8 



500 



UNIFIED MATHEMATICS 



Logarithms of numbers between 1000 and 1499. 








1 


2 


3 


4 


5 


6 


7 


8 


9 


100 


0000 


0004 


0009 


0013 


0017 


0022 


0026 


0030 


0035 


0039 


101 


0043 


0048 


0052 


0056 


0060 


0065 


0069 


0073 


0077 


0082 


102 


0086 


0090 


0095 


0099 


0103 


0107 


0111 


0116 


0120 


0124 


103 


0128 


0133 


0137 


0141 


0145 


0149 


0154 


0158 


0162 


0166 


104 


0170 


0175 


0179 


0183 


0187 


0191 


0195 


0199 


0204 


0208 


105 


0212 


0216 


0220 


0224 


0228 


0233 


0237 


0241 


0245 


0249 


106 


0253 


0257 


0261 


0265 


0269 


0273 


0278 


0282 


0286 


0290 


107 


0294 


0298 


0302 


0306 


0310 


0314 


0318 


0322 


0326 


0330 


108 


0334 


0338 


0342 


0346 


0350 


0354 


0358 


0362 


0366 


0370 


109 


0374 


0378 


0382 


0386 


0390 


0394 


0398 


0402 


0406 


0410 


110 


0414 


0418 


0422 


0426 


0430 


0434 


0438 


0441 


0445 


0449 


111 


0453 


0457 


0461 


0465 


0469 


0473 


0477 


0481 


0484 


0488 


112 


0492 


0496 


0500 


0504 


0508 


0512 


0515 


0519 


0523 


0527 


113 


0531 


0535 


0538 


0542 


0546 


0550 


0554 


0558 


0561 


0565 


114 


0569 


0573 


0577 


0580 


0584 


0588 


0592 


0596 


0599 


0603 


115 


0607 


0611 


0615 


0618 


0622 


0626 


0630 


0633 


0637 


0641 


116 


0645 


0648 


0652 


0656 


0660 


0663 


0667 


0671 


0674 


0678 


117 


0682 


0686 


0689 


0693 


0697 


0700 


0704 


0708 


0711 


0715 


118 


0719 


0722 


0726 


0730 


0734 


0737 


0741 


0745 


0748 


0752 


119 


0755 


0759 


0763 


0766 


0770 


0774 


0777 


0781 


0785 


0788 


120 


0792 


0795 


0799 


0803 


0806 


0810 


0813 


0817 


0821 


0824 


121 


0828 


0831 


0835 


0839 


0842 


0846 


0849 


0853 


0856 


0860 


122 


0864 


0867 


0871 


0874 


0878 


0881 


0885 


0888 


0892 


0896 


123 


0899 


0903 


0906 


0910 


0913 


0917 


0920 


0924 


0927 


0931 


124 


0934 


0938 


0941 


0945 


0948 


0952 


0955 


0959 


0962 


0966 







1 


2 


3 


4 


5 


6 ' 


7 


8 


9 


125 


0969 


0973 


0976 


0980 


0983 


0986 


0990 


0993 


0997 


1000 


126 


1004 


1007 


1011 


1014 


1017 


1021 


1024 


1028 


1031 


1035 


127 


1038 


1041 


1045 


1048 


1052 


1055 


1059 


1062 


1065 


1069 


128 


1072 


1075 


1079 


1082 


1086 


1089 


1092 


1096 


1099 


1103 


129 


1106 


1109 


1113 


1116 


1119 


1123 


1126 


1129 


1133 


1136 


130 


1139 


1143 


1146 


1149 


1153 


1156 


1159 


1163 


1166 


1169 


131 


1173 


1176 


1179 


1183 


1186 


1189 


1193 


1196 


1199 


1202 


132 


1206 


1209 


1212 


1216 


1219 


1222 


1225 


1229 


1232 


1235 


133 


1239 


1242 


1245 


1248 


1252 


1255 


1258 


1261 


1265 


1268 


134 


1271 


1274 


1278 


1281 


1284 


1287 


1290 


1294 


1297 


1300 


135 


1303 


1307 


1310 


1313 


1316 


1319 


1323 


1326 


1329 


1332 


136 


1335 


1339 


1342 


1345 


1348 


1351 


1355 


1358 


1361 


1364 


137 


1367 


1370 


1374 


1377 


1380 


1383 


1386 


1389 


1392 


1396 


138 


1399 


1402 


1405 


1408 


1411 


1414 


1418 


1421 


1424 


1427 


139 


1430 


1433 


1436 


1440 


1443 


1446 


1449 


1452 


1455 


1458 


140 


1461 


1464 


1467 


1471 


1474 


1477 


1480 


1483 


1486 


1489 


141 


1492 


1495 


1498 


1501 


1504 


1508 


1511 


1514 


1517 


1520 


142 


1523 


1526 


1529 


1532 


1535 


1538 


1541 


1544 


1547 


1550 


143 


1553 


1556 


1559 


1562 


1565 


1569 


1572 


1575 


1578 


1581 


144 


1584 


1587 


1590 


1593 


1596 


1599 


1602 


1605 


1608 


1611 


145 


1614 


1617 


1620 


1623 


1626 


1629 


1632 


1635 


1638 


1641 


146 


1644 


1647 


1649 


1652 


1655 


1658 


1661 


1664 


1667 


1670 


147 


1673 


1676 


1679 


1682 


1685 


1688 


1691 


1694 


1697 


1700 


148 


1703 


1706 


1708 


1711 ' 


1714 


1717 


1720 


1723 


1726 


1729 


149 


1732 


1735 


1738 


1741 


1744 


1746 


1749 


1752 


1755 


1758 







1 


2 


3 


4 


5 


6 


7 


8 


9 



TABLES 



501 



Logarithms of numbers between 1500 and 1999. 



150 


1761 


1764 


1767 


1770 


1772 


1775 


1778 


1781 


1784 


1787 


151 


1790 


1793 


1796 


1798 


1801 


1804 


1807 


1810 


1813 


1816 


152 


1818 


1821 


1824 


1827 


1830 


1833 


1836 


1838 


1841 


1844 


153 


1847 


1850 


1853 


1855 


1858 


1861 


1864 


1867 


1870 


1872 


151 


1875 


1878 


1881 


1884 


1886 


1889 


1892 


1895 


1898 


1901 


155 


1903 


1906 


1909 


1912 


1915 


1917 


1920 


1923 


1926 


1928 


156 


1931 


1934 


1937 


1940 


1942 


1945 


1948 


1951 


1953 


1956 


157 


1959 


1962 


1965 


1967 


1970 


1973 


1976 


1978 


1981 


1984 


158 


1987 


1989 


1992 


1995 


1998 


2000 


2003 


2006 


2009 


2011 


159 


2014 


2017 


2019 


2022 


2025 


2028 


2030 


2033 


2036 


2038 


160 


2041 


2044 


2047 


2049 


2052 


2055 


2057 


2060 


2063 


2066 


161 


2068 


2071 


2074 


2076 


2079 


2082 


2084 


2087 


2090 


2092 


162 


2095 


2098 


2101 


2103 


2106 


2109 


2111 


2114 


2117 


2119 


163 


2122 


2125 


2127 


2130 


2133 


2135 


2138 


2140 


2143 


2146 


164 


2148 


2151 


2154 


2156 


2159 


2162 


2164 


2167 


2170 


2172 


165 


2175 


2177 


2180 


2183 


2185 


2188 


2191 


2193 


2196 


2198 


166 


2201 


2204 


2206 


2209 


2212 


2214 


2217 


2219 


2222 


2225 


167 


2227 


2230 


2232 


2235 


2238 


2240 


2243 


2245 


2248 


2251 


168 


2253 


2256 


2258 


2261 


2263 


2266 


2269 


2271 


2274 


2276 


169 


2279 


2281 


2284 


2287 


2289 


2292 


2294 


2297 


2299 


2302 


170 


2304 


2307 


2310 


2312 


2315 


2317 


2320 


2322 


2325 


2327 


171 


2330 


2333 


2335 


2338 


2340 


2343 


2345 


2348 


2350 


2353 


172 


2355 


2358 


2360 


2363 


2365 


2368 


2370 


2373 


2375 


2378 


173 


2380 


2383 


2385 


2388 


2390 


2393 


2395 


2398 


2400 


2403 


174 


2405 


2408 


2410 


2413 


2415 


2418 


2420 


2423 


2425 


2428 







1 


2 


3 


4 


5 


6 


7 


8 


9 


175 


2430 


2433 


2435 


2438 


2440 


2443 


2445 


2448 


2450 


2453 


176 


2455 


2458 


2460 


2463 


2465 


2467 


2470 


2472 


2475 


2477 


177 


2480 


2482 


2485 


2487 


2490 


2492 


2494 


2497 


2499 


2502 


178 


2504 


2507 


2509 


2512 


2514 


2516 


2519 


2521 


2524 


2526 


179 


2529 


2531 


2533 


2536 


2538 


2541 


2543 


2545 


2548 


2550 


180 


2553 


2555 


2558 


2560 


2562 


2565 


2567 


2570 


2572 


2574 


181 


2577 


2579 


2582 


2584 


2586 


2589 


2591 


2594 


2596 


2598 


182 


2601 


2603 


2605 


2608 


2610 


2613 


2615 


2617 


2620 


2622 


183 


2625 


2627 


2629 


2632 


2634 


2836 


2639 


2641 


2643 


2646 


184 


2648 


2651 


2653 


2655 


2658 


2660 


2662 


2665 


2667 


2669 


185 


2672 


2674 


2676 


2679 


2681 


2683 


2686 


2688 


2690 


2693 


186 


2695 


2697 


2700 


2702 


2704 


2707 


2709 


2711 


2714 


2716 


187 


2718 


2721 


2723 


2725 


2728 


2730 


2732 


2735 


2737 


2739 


188 


2742 


2744 


2746 


2749 


2751 


2753 


2755 


2758 


2760 


2762 


189 


2765 


2767 


2769 


2772 


2774 


2776 


2778 


2781 


2783 


2785 


190 


2788 


2790 


2792 


2794 


2797 


2799 


2801 


2804 


2806 


2808 


191 


2810 


2813 


2815 


2817 


2819 


2822 


2824 


2826 


2828 


2831 


192 


2833 


2835 


2838 


2840 


2842 


2844 


2847 


2849 


2851 


2853 


193 


2856 


2858 


2860 


2862 


2865 


2867 


2869 


2871 


2874 


2876 


194 


2878 


2880 


2882 


2885 


2887 


2889 


2891 


2894 


2896 


2898 


195 


2900 


2903 


2905 


2907 


2909 


2911 


2914 


2916 


2918 


2920 


196 


2923 


2925 


2927 


2929 


2931 


2934 


2936 


2938 


2940 


2Q42 


197 


2945 


2947 


2949 


2951 


2953 


2956 


2958 


2960 


2962 


2964 


198 


2967 


2969 


2971 


2973 


2975 


2978 


2980 


2982 


2984 


2986 


199 


2989 


2991 


2993 


2995 


2997 


2999 


3002 


3004 


3006 


3008 







1 


2 


3 


4 


5 


6 


7 


8 


9 



502 



UNIFIED MATHEMATICS 



Log sin A° from 0° to 45°. 



A° 0' 



10' 



20' 



7.4637 7648 

1 8.2419 3088 3668 

2 5428 5776 6097 

3 7188 7423 7645 

4 8436 8613 8783 



30' 40' 



50' 



A° d. 



9408 *0658 *1627 *2419 89° 

4179 4637 5050 5428 88° 

6397 6677 6940 7188 87° 

7857 8059 8251 8436 86° 

8946 9104 9256 9403 85° 



Do not interpolate, but 
use the special table 
which gives these values 
by minutes. 



5 9403 

6 9.0192 

7 0859 

8 1436 

9 1943 

10 9.2397 

11 2806 

12 3179 

13 3521 

14 3837 

15 4130 

16 4403 

17 4659 

18 4900 

19 5126 

20 9.5341 

21 5543 

22 5736 

23 5919 

24 6093 



25 6259 

26 6418 

27 6570 

28 6716 

29 6856 

30 9.6990 

31 7118 

32 7242 

33 7361 

34 7476 

35 7586 

36 7692 

37 7795 

38 7893 

39 7989 

40 9.8081 

41 8169 
.42 8255 

43 8338 

44 8418 



60' 



9545 9682 
0311 0426 
0961 1060 
1525 1612 
2022 2100 

2468 2538 
2870 2934 
3238 3296 
3575 3629 
3887 3937 

4177 4223 
4447 4491 
4700 4741 
4939 4977 
5163 5199 

5375 5409 
5576 5609 
5767 5798 
5948 5978 
6121 6149 



6286 6313 
6444 6470 
6595 6620 
6740 6763 
6878 6901 

7012 7033 
7139 7160 
7262 7282 
7380 7400 
7494 7513 

7604 7622 
7710 7727 
7811 7828 
7910 7926 
8004 8020 

8096' 8111 
8184 8198 
8269 8283 
8351 8365 
8431 8444 

50' 40' 



9816 
0539 
1157 
1697 
2176 

2606 
2997 
3353 
3682 
3986 

4269 
4533 
4781 
5015 
5235 

5443 
5641 
5828 
6007 
6177 



6340 
6495 
6644 
6787 
6923 

7055 
7181 
7302 
7419 
7531 

7640 
7744 
7844 
7941 
8035 

8125 
8213 
8297 
8378 
8457 



9945 *0070 *0192 84° 
0648 0755 0859 83° 
1252 1345 1436 82° 91 
1781 1863 1943 81° 80 
2251 2324 2397 80° 73 



P. P. 



2674 2740 

3058 3119 

3410 3466 

3734 3786 

4035 4083 

4314 4359 

4576 4618 

4821 4861 

5052 5090 

5270 5306 

5477 5510 

5673 5704 

5859 5889 

6036 6065 

6205 6232 



6366 
6521 



6810 
6946 



7076 
7201 
7322 
7438 
7550 

7657 
7761 
7861 
7957 
8050 

8140 
8227 
8311 
8391 
8469 



6392 
6546 
6692 
6833 
6968 

7097 
7222 
7342 

7457 
7568 

7675 

7778 
7877 
7973 
8066 

8155 
8241 
8324 
8405 

8482 



2806 79° 
3179 78° 
3521 77° 
3837 76° 
4130 75° 



4403 74° 46 
4659 73° 43 
4900 72° 40 
5126 71° 38 
5341 70° 36 

5543 69° 34 
5736 68° 32 
5919 67° 31 
6093 66° 29 
6259 65° 28 



6418 64° 27 
6570 63° 25 
6716 62° 24 
6856 61° 23 
6990 60° 22 



7118 59° 21 
7242 58° 21 
7361 57° 20 
7476 56° 19 

7586 55° 18 

7692 54° 18 
7795 53° 17 
7893 52° 16 
7989 51° 16 
8081 50° 15 

8169 49° 15 
8255 48° 14 
8338 47° 14 
8418 46° 13 
8495 45° 13 





92 


90 88 


86 


84 


1 


9.2 


9.0 8.8 


8.6 


8.4 


2 


18.4 


18.0 17.6 


17.2 


16.8 


3 


27.6 


27.0 26.4 


25.8 


25.2 


4 


36.8 


36.0 35.2 


34.4 


33.6 


5 


46.0 


45.0 44.0 


43.0 


42.0 


6 


55.2 


54.0 52.8 


51.6 


50.4 


7 


64.4 


63.0 61.6 


60.2 


58.8 


8 


73.6 


72.0 70.4 


68.8 


67.2 


9 


82.8 


81.0 79.2 


77.4 


75.6 




82 


80 78 


76 


74 


1 


8.2 


8.0 7.8 


7.6 


7.4 


2 


16.4 


16.0 15.6 


15.2 


14.8 


3 


24.6 


24.0 23.4 


22.8 


22.2 


4 


32.8 


32.0 31.2 


30.4 


29.6 


5 


41.0 


40.0 39.0 


38.0 


37.0 


6 


49.2 


48.0 46.8 


45.6 


44.4 


7 


«7.4 


56.0 54.6 


53.2 


51.8 


8 


65.6 


64.0 62.4 


60.8 


59.2 


9 


73.8 


72.0 70.2 


68.4 


66.6 




72 


70 68 


66 


64 


1 


7.2 


7.0 6.8 


6.6 


6.4 


9 


14.4 


14.0 13.6 


13.2 


12.8 


3 


21.6 


21.0 20.4 


19.8 


19.2 


4 


28.8 


28.0 27.2 


26.4 


25.6 


5 


36.0 


35.0 34.0 


33.0 


32.0 


6 


43.2 


42.0 40.8 


39.6 


38.4 


7 


50.4 


49.0 47.6 


46.2 


44.8 



30' 



20' 10' 0' A d. 



8 57.6 56.0 54.4 52.8 51.2 

9 64.8 63.0 61.2 59.4 57.6 

62 60 58 56 54 

1 6.2 6.0 5.8 5.6 5.4 

2 12.4 12.0 11.6 11.2 10.8 

3 18.6 18.0 17.4 16.8 16.2 

4 24.8 24.0 23.2 22.4 21.6 

5 31.0 30.0 29.0 28.0 27.0 

6 37.2 36.0 34.8 33.6 32.4 

7 43.4 42.0 40.6 39.2 37.8 

8 49.6 48.0 46.4 44.8 43.2 

9 55.8 54.0 52.2 50.4 48.6 

52 50 48 46 44 

1 5.2 5.0 4.8 4.6 4.4 

2 10.4 10.0 9.6 9.2 8.8 

3 15.6 15.0 14.4 13.8 13.2 

4 20.8 20.0 19.2 18.4 17.6 

5 26.0 25.0 24.0 23.0 22.0 

6 31.2 30.0 28.8 27.6 26.4 

7 36.4 35.0 33.6 32.2 30.8 

8 41.6 40.0 38.4 36.8 35.2 

9 46.8 45.0 43.2 41.4 39.6 

P.P. 



Log cos A from 45° to 9G°. 



TABLES 



503 



Log sin A° from 45° to 90°, 



P.P. 



10 



20' 



30 



40' 50' 60' A° d. P. P. 





45 


8495 


8507 


8520 


8532 


8545 


8557 


8569 44 


12 




46 


8569 


85S2 


8594 


8606 


8618 


8629 


8641 


43 


12 


42 40 38 36 


47 


8641 


8653 


8665 


8676 


8688 


8699 


8711 


42 


12 


1 4.2 4.0 3.8 3.6 


48 


8711 


8722 


8733 


8745 


8756 


8767 


8778 41 


11 


2 8 4 8.0 7.6 7.2 


49 


8778 


8789 


8800 


8810 


8821 


8832 


8843 4( 


11 


3 12.6 12.0 11.4 10.8 






















4 16.8 16.0 15.2 14.4 






















5 21.0 20.0 19.0 18.0 


50 9.8* 


8853 


8864 


8874 


8884 


8895 


8905 


39 


10 


6 25.2 24.0 22.8 21.6 


51 


8905 


8915 


8925 


8935 


8945 


8955 


8965 


38 


10 


7 29.4 2S.0 26.6 25.2 


52 


8965 


8975 


8985 


8995 


9004 


9014 


9023 


37 


9 


8 33.6 32.0 30.4 28.8 

9 37.8 36.0 34.2 32.4 


53 


9023 


9033 


9042 


9052 


9061 


9070 


9080 36 


10 


54 


9080 


9089 


9098 


9107 


9116 


9125 


9134 


35 


9 


34 32 30 28 


55 


9134 


9142 


9151 


9160 


9169 


9177 


9186 34 


9 


56 


9186 


9194 


9203 


9211 


9219 


9228 


9236 33 


S 


1 3.4 3.2 3.0 2.8 

2 6.8 6.4 6.0 5.6 

3 10 2 96 90 84 


57 


9236 


9244 


9252 


9260 


9268 


9276 


9284 


32 


8 


58 


9284 


9292 


9300 


9308 


9315 


9323 


9331 


31 


8 


4 13.6 12.8 12.0 11.2 


59 


9331 


9338 


9£46 


9353 


9361 


9368 


9375 


30 


7 


5 17.0 16.0 15.0 14.0 






















6 20.4 19.2 1S.0 16.8 






















7 23.8 22.4 21.0 19.6 


60 9.9- 


8393 


9390 


9397 


9404 


9411 


9418 


29 


7 


8 27.2 25.6 24.0 22.4 


61 


9418 


9425 


9432 


9439 


9446 


9453 


9459 


28 


G 


9 30.6 28.8 27.0 25.2 


62 


9459 


9466 


9473 


9479 


9486 


9492 


9499 


27 


7 




63 


9499 


9505 


9512 


9518 


9524 


9530 


9537 


26 


7 




64 


9537 


9543 


9549 


9555 


9561 


9567 


9573 


25 


6 


26 24 22 20 






















1 2.6 2.4 2.2 2.0 


65 


9573 


9579 


9584 


9590 


9596 


9602 


9607 


24 


5 


2 5.2 4.8 4.4 4.0 


66 


9607 


9613 


9618 


9624 


9629 


9635 


9640 23 


5 


3 7.8 7.2 6.6 6.0 


67 


9640 


9646 


9651 


9656 


9661 


9667 


9672 


Ti 


5 


4 10.4 9.6 8.8 8.0 

5 13.0 12.0 11.0 10.0 

6 15.6 14.4 13.2 12.0 


68 
69 


9672 
9702 


9677 
9706 


9682 
9711 


9687 
9716 


9692 
9721 


9697 
9725 


9702 21 
9730 20 


5 
5 


7 18.2 16.8 15.4 14.0 






















8 20.8 19.2 17.6 16.0 










30' 












9 23.4 21.6 19.8 18.0 






















70 9.9730 


9734 


9739 


9743 


9748 


9752 


9757 


19 


5 




71 


9757 


9761 


9765 


9770 


9774 


9778 


9782 


18 


4 


19 18 17 16 


72 


9782 


9786 


9790 


9794 


9798 


9802 


9806 


17 


4 


1 1.9 1.8 1.7 1.6 


73 


9806 


9810 


9814 


9817 


9821 


9825 


9828 


16 


3 


2 3.8 3.6 3.4 3.2 


74 


9828 


9832 


9836 


9839 


9843 


9846 


9849 


15 


3 


3 5.7 5.4 5.1 4.8 






















4 7.6 7.2 6.8 6.4 






















5 9.5 9.0 8.5 8.0 


75 


9849 


9853 


9856 


9859 


9863 


9866 


9869 


14 


3 


6 11.4 10.8 10.2 9.6 


76 


9869 


9872 


9875 


9878 


9881 


9884 


9887 


13 


3 


7 13.3 12.6 11.9 11.2 


77 


9887 


9890 


9893 


9896 


9899 


9901 


9904 


12 


3 


8 15.2 14.4 13.6 12.8 

9 17.1 16.2 15.3 14.4 


78 
79 


9904 
9919 


9907 
9922 


9909 
9924 


9912 

9927 


9914 
9929 


9917 
9931 


9919 
9934 


11 

10 


2 
3 


15 14 13 12 


80 9.9934 


9936 


9938 


9940 


9942 


9944 


9946 


9 


2 


81 


9946 


9948 


9950 


9952 


9954 


9956 


9958 


8 


2 


1 1.5 1.4 1.3 1.2 

2 3.0 2.8 2.6 2.4 

3 4.5 4.2 3.9 3.6 

4 6.0 5.6 5.2 4.8 


82 


9958 


9959 


9961 


9963 


9964 


9966 


9968 


7 


2 


83 


9968 


9969 


9971 


9972 


9973 


9975 


9976 


a 


1 


84 


9976 


9977 


9979 


9980 


99X1 


9982 


9983 


5 


1 


5 7.5 7.0 6.5 6.0 






















6 9.0 8.4 7.8 7.2 

7 10.5 9.8 9.1 8.4 

8 12.0 11.2 10.4 9.6 


85 


9983 


9985 


9986 


9987 


9988 


9989 


9989 


4 





86 


9989 


9990 


9991 


9992 


9993 


9993 


9994 


3 


1 


■J 13.5 12.6 11.7 10.8 


87 


9994 


9995 


9995 


9996 


9996 


9997 


9997 


2 






9997 9998 9998 
9999 *0000 *0000 



P. P. 



60' 50' 



40' 



9999 9999 9999 9999 
*0000 *0000 *0000 *0000 



20' 10' 



12 

1.2 
2.4 
3.6 
4.8 
6.0 
7.2 
8.4 
9.6 
10.8 



Log cos A° from 0° to 45°. 



504 



UNIFIED MATHEMATICS 



Logarithms of tangents and cotangents, 0° to 23°. 



A° 0' 10' 30' 

O'tan 7.4637 7648 

log cot 2.5363 2352 

1° tan 8.2419 3089 3669 
log cot 1.7581 6911 6331 

2° tan 8.5431 5779 6101 
log cot 1.4569 4221 3899 

3° tan 8.7194 7429 7652 

log cot 1.2806 2571 2348 

4° tan 8.8446 8624 8795 

log cot 1.1554 1376 1205 



30' 



40' 50' 60' 



9409 *0658 *1627 *2419cot 89° 
0591 *9342 *8373 *7581 tan log 
4181 4638 5053 5431 cot 88° 
5819 5362 4947 4569 tan log 
6401 6682 6945 7194 cot 87° 
3599 3318 3055 2806 tan log 

7865 8067 8261 8446 cot 86° 

2135 1933 1739 1554 tan log 

8960 9118 9272 9420 cot 85° 

1040 0882 0728 0580 tan log 



5° tan 8.9420 


9563 9701 


9836 


9966*0093*0216 cot 84° 


log cot 1.0580 
6° tan 9.0216 


0437 0299 


0164 


0034 *9907 *9784 tan loir 


0336 0453 


0567 


0678 


0786 


0891 cot 83° 


log cot 0.9784 
7° tan 9.0891 


9664 9547 


9433 


9322 


9214 


9109 tan log 


0995 1096 


1194 


1291 


1385 


1478 cot 82° 


log cot 0.9109 


9005 8904 


8806 


8709 


8615 


8522 tan log 


8° tan 9. 1478 


1569 1658 


1745 


1831 


1915 


1997 cot 81° 


log cot 0.8522 
9° tan 9. 1997 


8431 8342 


8255 


8169 


8085 


8003 tan log 
2463 cot 80° 


2078 2158 


2236 


2313 


2389 


log cot 0.8003 


7922 7842 


7764 
30' 

2680 


7687 


7611 


7537 tan log 


10° tan 9.2463 


2536 2609 


2750 


2819 


2887 cot 79° 


log cot 0.7537 


7464 7391 


7320 


7250 


7181 


7113 tan log 
3275 cot 78° 


11° tan 9.2887 


2953 3020 


3085 


3149 


3212 


log cot 0.7113 


7047 6980 


6915 


6851 


6788 


6725 tan log 


12° tan 9.3275 


3336 3397 


3458 


3517 


3576 


3634 cot 77° 


log cot 0.6725 


6664 6603 


6542 


6483 


6424 


6366 tan log 


13° tan 9.3634 


3691 3748 


3804 


3859 


3914 


3968 cot 76° 


log cot 0.6366 


6309 6252 


6196 


6141 


6086 


6032 tan log 


14° tan 9.3968 


4021 4074 


4127 


4178 


4230 


4281 cot 75 


log cot 0.6032 


5979 5926 


5873 


5822 


5770 


5719 tan log 


15° tan 9.4281 


4331 4381 


4430 


4479 


4527 


4575 cot 74° 


log cot 0.5719 


5669 5619 


5570 


5521 


5473 


5425 tan log 


16° tan 9.4575 


4622 4669 


4716 


4762 


4808 


4853 cot 73° 


log cot 0.5425 


5378 5331 


5284 


5238 


5192 


5147 tan log 
5118 cot 72° 


17° tan 9.4853 


4898 4943 


4987 


5031 


5075 


log cot 0.5147 


5102 5057 


5013 


4969 


4925 


4882 tan log 


18° tan 9.5118 


5161 5203 


5245 


5287 


5329 


5370 cot 71° 


log cot 0.4882 


4839 4797 


4755 


4713 


4671 


4630 tan log 


19° tan 9.5370 


5411 5451 


5491 


5531 


5571 


5611 cot 70° 


log cot 0.4630 


4589 4549 


4509 


4469 


4429 


4389 tan log 


20° tan 9.5611 


5650 5689 


5727 


5766 


5804 


5842 cot 69° 


log cot 0.4389 


4350 4311 


4273 


4234 


4196 


4158 tan log 


21° tan 9.5842 


5879 5917 


5954 


5991 


6028 


6064 cot 68° 


log cot 0.4158 


4121 4083 


4046 


4009 


3972 


3936 tan log 


22° tan 9.6064 


6100 6136 


6172 


6208 


6243 


6279 cot 67° 


log cot 0.3936 


3900 3864 


3828 


3792 


3757 


3721 tan log 


A° 60' 


50' 40' 


30' 


20' 


10' 


0' A° 



Do not interpolate, 
but use the special 
table for log tan from 
0° to 9°, and log cot 
from 81° to 90°. 



P.P. 

82 80 78 76 74 

1 8.2 8.0 7.8 7.6 7.4 

2 16.4 16.015.6 15.2 14.8 

3 24.6 24.0 23.4 22.8 22.2 
432.8 32.031.2 30.4 29.6 
541.040 039.038.037.0 
6 49.2 48.0 46.8 45.6 44.4 

87 7 57.4 56.0 54.6 53.2 51.8 
87 8 65.6 64.0 62.4 60.8 59.2 
78 9 73.8 72.0 70.2 68.4 66.6 

78 72 70 68 66 64 

1 7.2 7.0 6.8 6.6 6.4 

2 14.4 14.013.6 13.2 12.8 
321.621.020.4 19.8 19.2 

4 28.8 28.0 27.2 26.4 25.6 
71 536.035.034.033.032.0 
71 6 43.2 42.0 40.8 39.6 38.4 
r > 7 50.4 49.0 47.6 46.2 44.8 



57.6 56.0 54.4 52.8 51.2 
64.8 63.0 61.2 59.4 57.6 



6 n 62 60 58 56 54 

1 6.2 6.0 5.8 5.6 5.4 

2 12.4 12.011.6 11.2 10.8 
56 3 18.6 18.017.4 16.8 16.2 
56 4 24.8 24.0 23.2 22.4 21.6 



52 



5 31.030.029.028.027.0 
co 6 37.2 36.0 34.8 33.6 32.4 
52 7 43.4 42.0 40.6 39.2 37.8 

8 49.6 48.0 46.4 44.8 43.2 



49 



9 55.8 54.0 52.2 50.4 48.6 



49 53 52 51 50 49 

46 1 5.3 5.2 5.1 5.0 4.9 
4 fi 2 10.6 10.4 10.2 10.0 9.8 
7a 3 15.9 15.6 15.3 15.014.7 
Tf 4 21.2 20.8 20.4 20.019.6 
44 5 26.5 26.0 25.5 25.0 24.5 
6 31.8 31.2 30.6 30.0 29.4 
, 7 37.136.4 35.7 35.0 34.3 
42 8 42.4 41.6 40.8 40.0 39.2 
42 9 47.7 46.8 45.9 45.0 44.1 
40 



40 



48 47 46 45 44 



1 4.8 4.7 4.6 4.5 4.4 

2 9.6 9.4 9.2 9.0 8.8 
39 3 14.4 14.113.813.513.2 
39 419.2 18.8 18.418.017.6 
37 5 24.0 23.5 23.0 22.5 22.0 
o, 6 28.8 28.2 27.6 27.026.4 
%' 7 33.6 32.9 32.2 31.5 30.8 
3 ° 838.4 37.636.836.035.2 
36 943.2 42.341.440.539.6 



.1 



P. P. 



Logarithms of tangents and cotangents, 67° to 90° 



TABLES 



505 



Logarithms of tangents and cotangents, 23° to 46°. 



P.P. 



10' 20' 30' 



40 



00' 



A° d. 







23° tan 9.6279 6314 6348 


6383 






log cot 0.3721 3686 3652 


3617 






24° tan 9.6486 6520 6553 


6587 


43 

1 4.3 


42 

4.2 


log cot 0.3514 3480 3447 


3413 


2 8.6 

3 12.9 


8.4 
12.6 


25° tan 9.6687 6720 6752 


6785 


4 17.2 


16.8 


log cot 0.3313 3280 3248 


3215 


5 21.5 


21.0 


26° tan 9.6882 6914 6946 


6977 


6 25.8 

7 30.1 

8 34 4 


25.2 
29.4 


log cot 0.3118 3086 3054 


3023 


33 6 


27° tan 9.7072 7103 7134 


7165 


9 38.7 


37.8 


log cot 0.2928 2897 2866 


2835 






28° tan 9.7257 7287 7317 


7348 


41 


40 


log cot 0.2743 2713 2683 


2652 


29° tan 9.7438 7467 7497 


7526 


1 4.1 

2 8 2 


4.0 
8 


log cot 0.2562 2533 2503 


2474 


3 12.3 


12.0 






4 16.4 


16.0 


30° tan 9.7614 7644 7673 


7701 


5 20.5 


20.0 


log cot 0.2386 2356 2327 


2299 


6 24.6 

7 28.7 

8 32.8 


24.0 
28.0 
32.0 


31° tan 9.7788 7816 7845 


7873 


log cot 0.2212 2184 2155 


2127 


9 36.9 


36.0 


32° tan 9.7958 7986 8014 


8042 






log cot 0.2042 2014 1986 


1958 


39 


38 


33° tan 9.8125 8153 8180 


8208 


1 3.9 

2 7 8 


3.8 

7 6 


log cot 0.1875 1847 1820 


1792 


34° tan 9.8290 8317 8344 


8371 


3 11.7 


11.4 


log cot 0.1710 1683 1656 


1629 


4 15.6 


11.4 






5 19.5 


19.0 




30' 


6 23.4 

7 27.3 

8 31.2 


22.8 
26.6 
30.4 




35° tan 9.8452 8479 8506 


8533 


9 35.1 


34.2 


log cot 0.1548 1521 1494 


1467 






36° tan 9.8613 8639 8666 


8692 


37 


36 


log cot 0.1387 1361 1334 


1308 


37° tan 9.8771 8797 8824 


8850 


1 3.7 

2 7.4 


3.6 

7.2 


log cot 0.1229 1203 1176 


1150 


3 11.1 


10.8 






4 14.8 


14.4 


38° tan 9.8928 8954 8980 


9006 


5 18.5 

6 22.2 

7 25 9 


18.0 
21.6 

25 2 


log cot 0.1072 1046 1020 


0994 


39° tan 9.9084 9110 9135 


9161 


8 29^6 


28^8 


log cot 0.0916 0890 0865 


0839 


9 33.3 


32.4 










40° tan 9.9238 9264 9289 


9315 






log cot 0.0762 0736 0711 


0685 


35 


34 


41° tan 9.9392 9417 9443 


9468 


log cot 0.0608 0583 0557 


0532 


1 3.5 

2 7.0 

3 10.5 


3.4 


42° tan 9.9544 9570 9595 


9621 


6.8 
10.2 


log cot 0.0456 0430 0405 


0379 


4 14.0 


13.6 






6 17.5 

6 21.0 

7 24.5 


17.0 
20.4 
23 8 


43° tan 9.9697 9722 9747 


9772 


log cot 0.0303 0278 0253 


0228 


8 28.0 


27.2 


44" tan 9.9848 9874 9899 


9924 


931.5 


30.6 


log cot 0.0152 0126 0101 


0076 






45° tan 0.0000 0025 0051 


0076 






log cot 0.0000*9975*9949 *9924 


P.P. 


A° 60' 50' 40' 


30' 



6417 6452 6486 cot 66° 34 
3583 3548 3514 tan log 34 
6620 6654 6687 cot 65° 33 
3380 3346 3313 tan log 33 

6817 6850 6882 cot 64° 33 
3183 3150 3118 tan log 33 
7009 7040 7072 cot 63° 32 
2991 2960 2928 tan log 32 
7196 7226 7257 cot 62° 31 
2804 2774 2743 tan log 31 

7378 7408 7438 cot 61° 30 
2622 2592 2562 tan log 30 
7556 7585 7614 cot 60° 29 
2444 2415 2386 tan log 29 

7730 7759 7788 cot 59° 29 
2270 2241 2212 tan log 29 
7902 7930 7958 cot 58° 28 
2098 2070 2042 tan log 28 
8070 8097 8125 cot 57° 28 
1930 1903 1875 tan log 28 

8235 8263 8290 cot 56° 27 
1765 1737 1710 tan log 27 
8398 8425 8452 cot 55° 27 
1602 1575 1548 tan log 27 



8559 8586 8613 cot 54° 27 
1441 1414 1387 tan log 27 
8718 8745 8771 cot 53° 26 
1282 1255 1229 tan log 26 
8876 8902 8928 cot 52° 26 
1124 1098 1072 tan log 26 

9032 9058 9084 cot 51° 26 
0968 0942 0916 tan log 26 
9187 9212 9238 cot 50° 26 
0813 0788 0762 tan log 26 

9341 9366 9392 cot 49° 26 
0659 0634 0608 tan log 26 
9494 9519 9544 cot 48° 25 
0506 0481 0456 tan log 25 
9646 9671 9697 cot 47° 25 
0354 0329 0303 tan log 25 

9798 9823 9848 cot 46° 25 
0202 0177 0152 tan log 25 
9949 9975 0000 cot 45° 25 
0051 0025 0000 tan log 25 
0101 0126 0152 cot 44° 25 
"9899*9874*9848 tan log 25 



P.P. 



33 32 



1 


3.3 


3.2 


2 


6.6 


6.4 


3 


9.9 


9.6 


4 


13.2 


12.8 


5 


16.5 


16.0 


6 


19.8 


19.2 


7 23.1 


22.4 


8 26.4 


25.6 


9 29.7 28.8 




31 


30 


1 


3.1 


3.0 


2 


6.2 


6.0 


3 


9.3 


9.0 


4 


12.4 


12.0 


5 


15.5 


15.0 


6 


18.6 


18.0 


7 


21.7 


21.0 


8 24.8 24.0 


9 27.9 


27.0 




29 


28 


1 


2.9 


2.8 


2 


5.8 


5.6 


3 


8.7 


8.4 


4 


11.6 


11.2 


5 


14.5 


14.0 


6 


17.4 


16.8 


7 


20.3 


19.6 


8 23.2 


22.4 


9 26.1 


25.2 




27 


26 


1 


2.7 


2.6 


2 


5.4 


5.2 


3 


8.1 


7.8 


4 


10.8 


10.4 


5 


13.5 


13.0 


6 


16.2 


15.6 


7 


18.9 


18.2 


8 21.6 20.8 


9 24.3 


23.4 




25 


24 


1 


2.5 


2.4 


2 


5.0 


4.8 


3 


7.5 


7.2 


4 


10.0 


9.6 


5 


12.5 


12.0 





15.0 


14.4 


7 


17.5 


16.8 


S 


20.0 


19.2 


') 


22.5 21.6 



20' 10' 



0' 



A° d. 



P. P. 



Logarithms of tangents and cotangents, 44° to 67°. 



506 UNIFIED MATHEMATICS 



Log sin by minutes from 0° to 9°. 

A ' 0' V 2' 3' 4' 5' 6' 7' 8' 9' 10' 

6.4637 7648 9408 *0658 *1627 *2419 *3088 *3668 *4180 *4637 50 

10 7.4637 5051 5429 5777 6099 6398 6678 6942 7190 7425 7648 40 



20 7648 


7859 8061 8255 


8439 


8617 


8787 


8951 


*109 


*261 


*408 30 


30 7.9408 


551 


689 


822 


952 


*078 


*200 


*319 


*435 


*548 


*658 20 


40 8.0658 


765 


870 


972 


*072 


*169 


*265 


*358 


*450 


*539 


*627 10 


50 8.1627 


713 


797 


880 


961 


*041 


*119 


*196 


*271 


*346 


*419 89 


1 8.2419 


490 


561 


630 


699 


766 


832 


898 


962 


*025 


*088 50 


10 8.3088 


150 


210 


270 


329 


388 


445 


502 


558 


613 


668 40 


20 668 


722 


775 


828 


880 


931 


982 


*032 


*082 


*131 


*179 30 


30 8.4179 


227 


275 


322 


368 


414 


459 


504 


549 


593 


637 20 


40 637 


680 


723 


765 


807 


848 


890 


930 


971 


*011 


*050 10 


50 8.5050 


090 


129 


167 


206 


243 


281 


318 


355 


392 


428 88 


2 8.5428 


464 


500 


535 


571 


605 


640 


674 


708 


742 


776 50 


10 776 


809 


842 


875 


907 


939 


972 


*003 


*035 


*066 


*097 40 


20 8.6097 


128 


159 


189 


220 


250 


279 


309 


339 


368 


397 30 


30 397 


426 


454 


483 


511 


539 


567 


595 


622 


650 


677 20 


40 677 


704 


731 


758 


784 


810 


837 


863 


889 


914 


940 10 


50 940 


965 


991 


*016 


*041 


*066 


*090 


*115 


*140 


*164 


*188 87 


3 8.7188 


212 


236 


260 


283 


307 


330 


354 


377 


400 


423 50 


10 423 


445 


468 


491 


513 


535 


557 


580 


602 


623 


645 40 


20 645 


667 


688 


710 


731 


752 


773 


794 


815 


836 


857 30 


30 857 


877 


898 


918 


939 


959 


979 


999 


*019 


*039 


*059 20 


40 8.8059 


078 


098 


117 


137 


156 


175 


194 


213 


232 


251 10 


50 251 


270 


289 


307 


326 


345 


363 


381 


400 


418 


436 86 


4 8.8436 


454 


472 


490 


508 


525 


543 


560 


578 


595 


613 50 


10 613 


630 


647 


665 


682 


699 


716 


733 


749 


766 


783 40 


20 783 


799 


816 


833 


849 


865 


882 


898 


914 


930 


946 30 


30 946 


962 


978 


994 


*010 


*026 


*042 


*057 


*£73 


*089 


*104 20 


40 8.9104 


119 


135 


150 


166 


181 


196 


211 


226 


241 


256 10 


50 256 


271 


286 


301 


315 


330 


345 


359 


374 


389 


403 85 


5 8.9403 


417 


432 


446 


460 


475 


489 


503 


517 


531 


545 50 


10 545 


559 


573 


587 


601 


614 


628 


642 


655 


669 


682 40 


20 682 


696 


709 


723 


736 


750 


763 


776 


789 


803 


816 30 


30 816 


829 


842 


855 


868 


881 


894 


907 


919 


932 


945 20 


40 945 


958 


970 


983 


996 


*008 


*021 


*033 


*046 


*058 


*070 10 


50 9.0070 


083 


095 


107 


120 


132 


144 


156 


168 


180 


192 84 


6 9.0192 


204 


216 


228 


240 


252 


264 


276 


287 


299 


311 50 


10 311 


323 


334 


346 


357 


369 


380 


392 


403 


415 


426 40 


20 426 


438 


449 


460 


472 


483 


494 


505 


516 


527 


539 30 


30 539 


550 


561 


572 


583 


594 


605 


616 


626 


637 


648 20 


40 648 


659 


670 


680 


691 


702 


712 


723 


734 


744 


755 10 


50 755 


765 


776 


786 


797 


807 


818 


828 


838 


849 


859 83 


7 9.0859 


869 


879 


890 


900 


910 


920 


930 


940 


951 


961 50 


10 961 


971 


981 


991 


*001 


*011 


*020 


*030 


*040 


*050 


*060 40 


20 9.1060 


070 


080 


089 


099 


109 


118 


128 


138 


147 


157 30 


30 157 


167 


176 


186 


195 


205 


214 


224 


233 


242 


252 20 


40 252 


261 


271 


280 


289 


299 


308 


317 


326 


336 


345 10 


50 345 


354 


363 


372 


381 


390 


399 


409 


418 


427 


436 82 


8 9.1436 


445 


453 


462 


471 


480 


489 


498 


507 


510 


525 50 


10 525 


533 


542 


551 


560 


568 


577 


586 


594 


603 


612 40 


20 612 


620 


629 


637 


646 


655 


663 


672 


680 


689 


697 30 


30 697 


705 


714 


722 


731 


739 


747 


756 


764 


772 


781 20 


40 781 


789 


797 


806 


814 


822 


830 


838 


847 


855 


863 10 


50 863 


871 


879 


887 


895 


903 


911 


919 


927 


935 


943 81 


A 10' 


9' 


8' 


7' 


6' 


5' 


4' 


3' 


2' 


1' 


0' ' A 



Log cos by minutes from 81° to 90°. 



TABLES 



507 



Log tan by minutes from 0° to 9°. 



10' 






6.4637 7648 940S *0658 


*1627 


*2419 *3088 *3668 *41S0 *4637 50 


10 7.4637 


5051 5429 5777 


6099 


6398 


6678 


6942 


7190 


7425 


7648 40 


20 7648 


7860 8062 8255 


8439 


8617 


8787 


8951 


*109 


*261 


*409 30 


30 7.9409 


551 


689 


823 


952 


*078 


*200 


*319 


*435 


*548 


*658 20 


40 8.0658 


765 


870 


972 


*072 


*170 


*265 


*359 


*450 


*540 


*627 10 


50 8.1627 


713 


798 


880 


962 


*041 


*120 


*196 


*272 


*346 


*419 89 


1 8.2419 


491 


562 


631 


700 


767 


833 


899 


963 


*026 


*089 50 


10 S.3089 


150 


211 


271 


330 


389 


446 


503 


559 


614 


669 40 


20 669 


723 


776 


829 


881 


932 


983 


*033 


*083 


132 


181 30 


30 8.4181 


229 


276 


323 


370 


416 


461 


506 


551 


595 


638 20 


40 638 


682 


725 


767 


809 


851 


892 


933 


973 


*013 


*053 10 


50 8.5053 


092 


131 


170 


208 


246 


283 


321 


358 


394 


431 88 


2 8.5431 


467 


503 


538 


573 


608 


643 


677 


711 


745 


779 50 


10 779 


812 


845 


878 


911 


943 


975 


*007 


*038 


*070 


♦101 40 


20 8.6101 


132 


163 


193 


223 


254 


283 


313 


343 


372 


401 30 


30 401 


430 


459 


487 


515 


; 544 


571 


599 


627 


654 


682 20 


40 682 


709 


736 


762 


789 


815 


842 


868 


894 


920 


945 10 


50 945 


971 


996 *021 


*046 


*071 


*096 


*121 


*145 


*170 


*194 87 


3 8.7194 


218 


242 


266 


290 


313 


337 


360 


383 


406 


429 50 


10 429 


452 


475 


497 


520 


542 


565 


587 


609 


631 


652 40 


20 652 


674 


696 


717 


739 


760 


781 


802 


823 


844 


865 30 


30 865 


886 


906 


927 


947 


967 


988 


*008 


*028 


*048 


*067 20 


40 8.8067 


087 


107 


126 


146 


165 


185 


204 


223 


242 


261 10 


50 261 


280 


299 


317 


336 


355 


373 


392 


410 


428 


446 86 


4 8.8446 


465 


483 


501 


518 


536 


554 


572 


589 


607 


624 50 


10 624 


642 


659 


676 


694 


711 


728 


745 


762 


778 


795 40 


20 795 


812 


829 


845 


862 


878 


895 


911 


927 


944 


960 30 


30 960 


976 


992 *008 


*024 


*040 


*056 


*071 


*087 


*103 


*118 20 


40 8.9118 


134 


150 


165 


180 


196 


211 


226 


241 


256 


272 10 


50 272 


287 


302 


316 


331 


346 


361 


376 


390 


405 


420 85 


5 8.9420 


434 


449 


463 


477 


492 


506 


520 


534 


549 


563 50 


10 563 


577 


591 


605 


619 


633 


646 


660 


674 


688 


701 40 


20 701 


715 


729 


742 


756 


769 


782 


796 


809 


823 


836 30 


30 836 


849 


862 


875 


888 


901 


915 


927 


940 


953 


966 20 


40 966 


979 


992 *005 


*017 


*030 


*043 


*055 


*068 


*080 


*093 10 


50 9.0093 


105 


118 


130 


143 


155 


167 


180 


192 


204 


216 84 


6 9.0216 


228 


240 


253 


265 


277 


289 


300 


312 


324 


336 50 


10 336 


348 


360 


371 


383 


395 


407 


418 


430 


441 


453 40 


20 453 


464 


476 


487 


499 


510 


521 


533 


544 


555 


567 30 


30 567 


578 


589 


600 


611 


622 


633 


645 


656 


667 


678 20 


40 678 


688 


699 


710 


721 


732 


743 


754 


764 


775 


786 10 


50 786 


796 


807 


818 


828 


839 


849 


860 


871 


881 


891 83 


7 9.0891 


902 


912 


923 


933 


943 


954 


964 


974 


984 


995 50 


10 995 


*005 


*015 


*025 


*035 


*045 


*055 


*066 


*076 


*086 


*096 40 


20 9.1096 


106 


116 


125 


135 


145 


155 


165 


175 


185 


194 30 


30 194 


204 


214 


223 


233 


243 


252 


262 


272 


281 


291 20 


40 291 


300 


310 


319 


329 


338 


348 


357 


367 


376 


385 10 


50 385 


395 


404 


413 


423 


432 


441 


450 


460 


469 


478 82 


8 9.1478 


487 


496 


505 


515 


524 


533 


542 


551 


560 


569 50 


10 569 


578 


587 


596 


605 


613 


622 


631 


640 


649 


658 40 


20 658 


667 


675 


684 


693 


702 


710 


719 


728 


736 


745 30 


30 745 


754 


762 


771 


779 


788 


797 


805 


814 


822 


831 20 


40 831 


839 


848 


856 


864 


873 


881 


890 


898 


906 


915 10 


50 915 


923 


931 


940 


948 


956 


964 


973 


981 


989 


997 81 



10' 



Log cot by minutes from 81° to 90°. 



508 



UNIFIED MATHEMATICS 



Numerical values of the sine function, 0° to 45°. 



10' 20' 



30' 



10' 



50' 



P.P. 






0.0000 


0029 


0058 


0087 


0116 


0145 


0175 


89 


29 








1 


0175 


0204 


0233 


0262 


0291 


0320 


0349 


88 


29 








2 


0349 


0378 


0407 


0436 


0465 


0494 


0523 


87 


29 




30 


29 


3 


0523 


0552 


0581 


0610 


0640 


0669 


0698 


86 


29 


1 


3.0 

6.0 


2 9 


4 


0698 


0727 


0756 


0785 


0814 


0843 


0872 


85 


29 


2 


5^8 






















3 

4 
5 


9.0 
12.0 
15.0 


8.7 
11.6 
14.5 


5 


0.0872 


0901 


0929 


0958 


0987 


1016 


1045 


84 


29 


6 


1045 


1074 


1103 


1132 


1161 


1190 


1219 


83 


29 


6 


18.0 


17.4 


7 


1219 


1248 


1276 


1305 


1334 


1363 


1392 


82 


29 


7 


21.0 


20.3 


8 


1392 


1421 


1449 


1478 


1507 


1536 


1564 


81 


29 


8 
9 


24.0 
27.0 


23.2 
26.1 


9 


1564 


1593 


1622 


1650 


1679 


1708 


1736 


80 


29 


10 


0.1736 


1765 


1794 


1822 


1851 


1880 


1908 


79 


29 




28 


27 


11 


1908 


1937 


1965 


1994 


2022 


2051 


2079 


78 


28 




12 


2079 


2108 


2136 


2164 


2193 


2221 


2250 


77 


28 


1 
2 
3 


2.8 
5.6 
8.4 


2.7 
5.4 
8.1 


13 


2250 


2278 


2306 


2334 


2363 


2391 


2419 


76 


28 


14 


2419 


2447 


2476 


2504 


2532 


2560 


2588 


75 


28 


4 
5 

6 

7 


11.2 
14.0 
16.8 
19.6 


10.8 
13.5 
16.2 
18.9 


15 


0.2588 


2616 


2644 


2672 


2700 


2728 


2756 


74 


28 


16 


2756 


2784 


2812 


2840 


2868 


2896 


2924 


73 


28 


8 


22.4 


21.3 


17 


2924 


2952 


2979 


3007 


3035 


3062 


3090 


72 


28 


9 


25.2 


24.3 


18 


3090 


3118 


3145 


3173 


3201 


3228 


3256 


71 


28 








19 


3256 


3283 


3311 


3338 


3365 


3393 


3420 


70 


27 




26 


25 


20 


0.3420 


3448 


3475 


3502 


3529 


3557 


3584 


69 


27 


1 


2.6 


2.5 


21 


3584 


3611 


3638 


3665 


3692 


3719 


3746 


68 


27 


2 


5.2 

7.8 

10.4 

13.0 


5.0 
7.5 
10.0 
12.5 


22 


3746 


3773 


3800 


3827 


3854 


3881 


3907 


67 


27 


3 

4 
5 


23 


3907 


3934 


3961 


3987 


4014 


4041 


4067 


66 


* 27 


21 


4067 


4094 


4120 


4147 
30' 


4173 


4200 


4226 


65 


26 


6 
7 
8 
9 


15.6 
18.2 
20.8 
23.4 


15.0 
17.5 
20.0 
22.5 


25 


0.4226 


4253 


4279 


4305 


4331 


4358 


4384 


64 


26 








26 


4384 


4410 


4436 


4462 


4488 


4514 


4540 


63 


26 








27 


4540 


4566 


4592 


4617 


4643 


4669 


4695 


62 


26 




24 


23 


28 


4695 


4720 


4746 


4772 


4797 


4823 


4848 


61 


26 


1 


2.4 


2.3 


29 


4848 


4874 


4899 


4924 


4950 


4975 


5000 


60 


25 


2 
3 
4 
5 


4.8 

7.2 

9.6 

12.0 


4.6 

6.9 

9.2 

11.5 


30 


0.5000 


5025 


5050 


5075 


5100 


5125 


5150 


59 


25 


31 


5150 


5175 


5200 


5225 


5250 


5275 


5299 


58 


25 


6 


14.4 


13.8 


32 


5299 


5324 


5348 


5373 


5398 


5422 


5446 


57 


24 


7 
8 
9 


16.8 
19.2 
21.6 


16.1 
18.4 
20.7 


33 


5446 


5471 


5495 


5519 


5544 


5568 


5592 


56 


24 


31 


5592 


5616 


5640 


5664 


5688 


5712 


5736 


55 


24 


35 


0.5736 


5760 


5783 


5807 


5831 


5854 


5878 


54 


24 


29! 91 2A 


36 


5878 


5901 


5925 


5948 


5972 


5995 


6018 


53 


23 


1 2 

2 4 

3 6 


.2 2.1 2.0 
4 49 4. n 


37 


6018 


6041 


6065 


6088 


6111 


6134 


6157 


52 


23 


38 


6157 


6180 


6202 


6225 


6248 


6271 


6293 


51 


23 


16 6 


.3 6.0 


39 


6293 


6316 


6338 


6361 


6383 


6406 


6428 


50 


22 


4 8.8 8.4 8.0 

5 11.0 10.5 10.0 

6 13.2 12.6 12.0 


10 


0.6428 


6450 


6472 


6494 


6517 


6539 


6561 


49 


22 


7 15.4 14.7 14.0 


41 


6561 


6583 


6604 


6626 


6648 


6670 


6691 


48 


22 


8 17.6 16.8 16.0 


42 


6691 


6713 


6734 


6756 


6777 


6799 


6820 


47 


22 


9 19.8 18.9 18.0 


43 


6820 


6841 


6862 


6884 


6905 


6926 


6947 


46 


21 








44 


6947 


6967 


6988 


7009 


7030 


7050 


7071 


45 


21 









50' 40' 



30' 



20' 10' 



P.P. 



Numerical values of the cosine function, 45° to 90°. 



TABLES 



509 



Numerical values of the sine function, 45° to 90°. 



0' 10' 20' 



30' 



40' 50' 



P. P. 



45 


0.7071 


7092 


7112 


7133 


7153 


7173 


7193 


44 


20 












it; 


7193 


7214 


7234 


7254 


7274 


7294 


7314 


43 


20 




21 


20 


19 


18 


47 


7314 


7333 


7353 


7373 


7392 


7412 


7431 


42 


20 


1 


2.1 


2.0 


1.9 


1.8 


4S 


7431 


7451 


7470 


7490 


7509 


7528 


7547 


41 


19 


2 


4.2 


4.0 


3.8 


3.6 


49 


7.i47 


7566 


7585 


7604 


7623 


7642 


7660 


40 


19 


3 
4 
5 


6.3 
8.4 
10.5 


6.0 
8.0 
10.0 


5.7 
7.6 
9.5 


5.4 
7.2 
9.0 






















50 


0.7660 


7679 


7698 


7716 


7735 


7753 


7771 


39 


18 


6 


12.6 


12.0 


11.4 


10.8 


51 


7771 


7790 


7808 


7826 


7844 


7862 


7880 


38 


18 


7 
8 
g 


14.7 
16.8 
18.9 


14.0 
16.0 
18.0 


13.3 
15.2 
17.1 


12.6 
14.4 
16.2 


52 


7SS0 


7898 


7916 


7934 


7951 


7969 


7986 


37 


18 


53 


7986 


S004 


8021 


8039 


8056 


8073 


8090 


36 


17 










54 


S090 


8107 


8124 


8141 


8158 


8175 


8192 


35 


17 




17 


16 


15 


14 


55 


0.8192 


8208 


8225 


8241 


8258 


8274 


8290 


34 


16 


1 


1.7 


1.6 


1.5 


1.4 


56 


8290 


8307 


8323 


8339 


8355 


8371 


8387 


33 


16 


2 
3 
4 


3.4 
5.1 

6 8 


3.2 
4.7 
6 4 


3.0 
4.5 
6 


2.8 
4.2 
5 6 


57 


8387 


S403 


8418 


8434 


8450 


8465 


8480 


32 


16 


5S 


8480 


8496 


8511 


8526 


8542 


8557 


8572 


31 


15 


5 


8.5 


8.0 


7.5 


7.0 


59 


8572 


8587 


8601 


8616 


8631 


8646 


8660 


30 


15 


6 

7 

8 


10.2 
11.9 
13.6 


9.6 
11.2 
12.8 


9.0 
10.5 
12.0 


8.4 
9.8 
11.2 


HO 


0.8660 


8675 


8689 


8704 


8718 


8732 


8746 


29 


14 


9 


15.3 


14.4 


13.5 


12.6 


61 


8746 


8760 


8774 


8780 


8802 


8816 


8829 


28 


14 












62 


8829 


8843 


8857 


8870 


8884 


8897 


8910 


27 


14 












63 


8910 


8923 


8936 


8949 


8962 


8975 


8988 


26 


13 




13 


12 


11 


10 


64 


8988 


9001 


9013 


9026 


9038 


9051 


9063 


25 


12 


1 

2 
3 


1.3 

2.6 
3.9 


1.2 
2.4 
3.6 


1.1 

2.2 
3.3 


1.0 
2.0 
3.0 


65 


0.9063 


9075 


9088 


9100 


9112 


9124 


9135 


24 


12 


4 


5.2 


4.8 


4.4 


4.0 


66 


9135 


9147 


9159 


9171 


9182 


9194 


9205 


23 


12 


5 


6.5 


6.0 


5.5 


5.0 


67 


9205 


9216 


9228 


9239 


9250 


9261 


9272 


22 


11 


6 
7 
8 


7.8 
9.1 
10 4 


7.2 
8.4 
9.6 


6.6 

7.7 
8.8 


6.0 
7.0 
8.0 


68 


9272 


9283 


9293 


9304 


9315 


9325 


9336 


21 


11 


69 


9336 


9346 


9356 


9367 
30' 

9426 


9377 


9387 


9397 


20 


10 


9 


11.7 


10.8 


9.9 


9.0 


70 


0.9397 


9407 


9417 


9436 


9446 


9455 


19 


10 












71 


9455 


9465 


9474 


9483 


9492 


9502 


9511 


18 


9 












rz 


9511 


9520 


9528 


9537 


9546 


9555 


9563 


17 


9 












73 


9563 


9572 


9580 


9588 


9596 


9605 


9613 


16 


8 












74 


9613 


9621 


9628 


9636 


9644 


9652 


9659 


15 


8 












75 


0.9659 


9667 


9674 


9681 


9689 


9696 


9703 


14 


7 












76 


9703 


9710 


9717 


9724 


9730 


9737 


9744 


13 


7 












77 


9744 


9750 


9757 


9763 


9769 


9775 


9781 


12 


6 












78 


9781 


9787 


9793 


9799 


9805 


9811 


9816 


11 


6 












79 


9816 


9822 


9827 


9833 


9838 


9843 


9848 


10 


5 




Interpolate men- 






















tally, using the mul- 


SO 


0.9848 


9853 


9858 


9863 


9868 


9872 


9877 


9 


5 


tiplication table. 




81 


9877 


9881 


9886 


9890 


9894 


9899 


9903 


8 


4 












82 


9903 


9907 


9911 


9914 


9918 


9922 


9925 


7 


4 












83 


9925 


9929 


9932 


9936 


9939 


9942 


9945 


6 


3 












84 


9945 


9948 


9951 


9954 


9957 


9959 


9962 


5 


3 












85 


0.9962 


9964 


9967 


9969 


9971 


9974 


9976 


4 


2 












86 


9976 


9978 


9980 


9981 


9983 


9985 


9986 


3 


2 












S7 


9986 


9988 


9989 


9990 


9992 


9993 


9994 


2 


1 












88 


9994 


9995 


9996 


9997 


9997 


9998 


9998 


1 


1 












89 


9998 


9999 


9999 


*0000 


*oooo 


*0000 


♦0000 



















60' 50' 40' 



30' 



•JO' 



P. P. 



Numerical values of the cosine function, 0° to 45 c 



510 



UNIFIED MATHEMATICS 



Numerical values of the tangent function, 0° to 45 c 



° 0' 10' 20' 

0.0000 0029 0058 

1 0175 0204 0233 

2 0349 0378 0407 

3 0524 0553 0582 

4 0699 0729 0758 

5 0.0875 0904 0934 

6 1051 1080 1110 

7 1228 1257 1287 

8 1405 1435 1465 

9 1584 1614 1644 

10 0.1763 1793 1823 

11 1944 1974 2004 

12 2126 2156 2186 

13 2309 2339 2370 

14 2493 2524 2555 



15 

i« 

17 
18 
19 

20 

21 

2'i 
83 
24 



30 
31 
32 
33 



35 



0.2679 2711 2742 

2867 2899 2931 

3057 3089 3121 

3249 3281 3314 

3443 3476 3508 

0.3640 3673 3706 

3839 3872 3906 

4040 4074 4108 

4245 4279 4314 

4452 4487 4522 



25 0.4663 4699 4734 

26 4877 4913 4950 

27 5095 5132 5169 

28 5317 5354 5392 

29 5543 5581 5619 



0.5774 5812 5851 

6009 6048 6088 

6249 6289 6330 

6494 6536 6577 

6745 6787 6830 



0.7002 7046 7089 

36 7265 7310 7355 

37 7536 7581 7627 

38 7813 7860 7907 

39 8098 8146 8195 

40 0.8391 8441 8491 

41 8693 8744 8796 

42 9004 9057 9110 

43 9325 9380 9435 

44 9657 9713 9770 

60' 50' 40' 



30' 

0087 
0262 
0437 
0612 

0787 

0963 
1139 
1317 
1495 
1673 

1853 
2035 
2217 
2401 
2586 

2773 
3962 
3153 
3346 
3541 

3739 
3939 
4142 
4348 
4557 

30' 

4770 
4986 
5206 
5430 
5658 

5890 
6128 
6371 
6619 
6873 

7133 
7400 
7673 
7954 
8243 

8541 
8847 
9163 
9490 

9827 



40' 50' 60' 



0116 0145 

0291 0320 

0466 0495 

0641 0670 

0816 0846 

0992 1022 

1169 1198 

1346 1376 

1524 1554 

1703 1733 

1883 1914 

2065 2095 

2247 2278 

2432 2462 

2617 2648 

2805 2836 

2994 3026 

3185 3217 

3378 3411 

3574 3607 

3772 3805 

3973 4006 

4176 4210 

4383 4417 

4592 4628 



0175 89 29 

0349 88 29 

0524 87 29 

0699 86 29 

0875 85 29 



1051 
1228 
1405 
1584 
1763 

1944 
2126 
2309 
2493 
2679 

2867 
3057 
3249 
3443 
3640 

3839 
4040 
4245 
4452 
4663 



4806 4841 4877 

5022 5059 5095 

5243 5280 5317 

5467 5505 5543 

5696 5735 5774 

5930 5969 6009 

6168 6208 6249 

6412 6453 6494 

6661 6703 6745 

6916 6959 7002 



84 29 

83 30 

82 30 

81 30 

80 30 

79 30 

78 30 

77 30 

76 31 

75 31 

74 31 

73 32 

72 32 



69 33 

68 34 

67 34 

66 34 

65 35 



64 36 

63 36 

62 37 

61 38 

60 38 

59 39 

58 40 

57 41 

56 42 

55 43 



7177 7221 7265 54 44 

7445 7490 7536 53 45 

7720 7766 7813 52 46 

8002 8050 8098 51 48 

8292 8342 8391 50 49 

8591 8642 8693 49 50 

8899 8952 9004 48 52 

9217 9271 9325 47 54 

9545 9601 9657 46 55 

9884 9942 *0000 45 57 



P. P. 

29 30 31 32 33 

1 2.9 3.0 3.1 3.2 3.3 

2 5.8 6.0 6.2 6.4 6.6 

3 8.7 9.0 9.3 9.6 9.9 

4 11.6 12.0 12.4 12.8 13.2 

5 14.5 15.0 15.5 16.0 16.5 

6 17.4 18.0 18.6 19.2 19.8 

7 20.3 21.0 21.7 22.4 23.1 

8 23.2 24.0 24.8 25.6 26.4 

9 26.1 27.0 27.9 28.8 29.7 

34 35 36 37 38 

1 3.4 3.5 3.6 3.7 3.8 

2 6.8 7.0 7.2 7.4 7.6 

3 10.2 10.5 10.8 11.1 11.4 

4 13.6 14.0 14.4 14.8 15.2 

5 17.0 17.5 18.0 18.5 19.0 

6 20.4 21.0 21.6 22.2 22.8 

7 23.8 24.5 25.2 25.9 26.6 

8 27.2 28.0 28.8 29.6 30.4 

9 30.6 31.5 32.4 33.3 34.2 

39 40 41 42 43 

1 3.9 4.0 4.1 4.2 4.3 

2 7.8 8.0 8.2 8.4 8.6 

3 11.7 12.0 12.3 12.6 12.9 

4 15.6 16.0 16.4 16.8 17.2 

5 19.5 20.0 20.5 21.0 21.5 

6 23.4 24.0 24.6 25.2 25.8 

7 27.3 28.0 28.7 29.4 30.1 

8 31.2 32.0 32.8 33.6 34.4 

9 35.1 36.0 36.9 37.8 38.7 



44 

4.4 



45 



20' 



10' 



0' 



46 47 48 

1 4.4 4.5 4.6 4.7 4.8 

2 8.8 9.0 9.2 9.4 9.6 

3 13.2 13.5 13.8 14.1 14.4 

4 17.6 18.0 18.4 18.8 19.2 

5 22.0 22.5 23.0 23.5 24.0 

6 26.4 27.0 27.6 28.2 28.8 

7 30.8 31.5 32.2 32.9 33.6 

8 35.2 36.0 36.8 37.6 38.4 

9 39.6 40.5 41.4 42.3 43.2 

49 50 51 52 53 

1 4.9 5.0 5.1 5.2 5.3 

2 9.8 10.0 10.2 10.4 10.6 

3 14.7 15.0 15.3 15.6 15.9 

4 19.6 20.0 20.4 20.8 21.2 

5 24.5 25.0 25.5 26.0 26.5 

6 29.4 30.0 30.6 31.2 31.8 

7 34.3 35.0 35.7 36.4 37.1 

8 39.2 40.0 40.8 41.6 42.4 

9 44.1 45.0 45.9 46.8 47.7 

54 55 56 57 58 

1 5.4 5.5 5.6 5.7 5.8 

2 10.8 11.0 11.2 11.4 11.6 

3 16.2 16.5 16.8 17.1 17.4 

4 21.6 22.0 22.4 22.8 23.2 

5 27.0 27.5 28.0 28.5 29.0 

6 32.4 33.0 33.6 34.2 34.8 

7 37.8 38.5 39.2 39.9 40.6 

8 43.2 44.0 44.8 45.6 46.4 

9 48.6 49.5 50.4 51.3 52.2 

P. P. 



Numerical values of the cotangent function, 45° to 90 c 



TABLES 511 

Numerical values of the tangent function, 45° to 90°. 

° 0' 10' 20' 30' 40' 50' 60' d. P. P. 

45 1.000 1.006 1.012 LOIS 1.024 1.030 1.036 44 6 6 7 8 9 10 11 

46 1.036 1.042 1.04S 1.054 1.060 1.066 1.072 43 6 1060708091011 

47 1.072 1.079 1.085 1.091 1.098 1.104 1.11143 6 2 L2 li L6 U 2lo 2.2 

48 1.111 1.117 1.124 1.130 1.137 1.144 1.150 41 6 3 1.8 2.12.4 2.7 3.0 3.3 

49 1.150 1.157 1.164 1.171 1.178 1.185 1.192 40 7 4 2.4 2.8 3.2 3.6 4.0 4.4 

5 3.0 3.o 4.0 4.5 5.0 5.5 

6 3.6 4.2 4.8 5.4 6.0 6.6 

50 1.192 1.199 1.206 1.213 1.220 1.228 1.235 39 7 7 4.2 4.9 5.6 6.3 7 .0 7.7 

51 1.235 1.242 1.250 1.257 1.265 1.272 1.280 38 8 q H «, S, « nn oo 
53 1.2S0 1.2S8 1.295 1.303 1.311 1.319 1.327 37 8 »&•* «*•*'•* a-* »■" "■» 

53 1.327 1.335 1.343 1.351 1.360 1.368 1.376 36 8 i* i« «a i« i« 

54 1.376 1.385 1.393 1.402 1.411 1.419 1.428 35 9 , x " Ai x * xo J * 

1 1.2 1.3 1.4 1.5 1.6 

9 9 i 9 A 9 52 ^n *\ 9 

55 1.42S 1.437 1.446 1.455 1.464 1.473 1.483 34 9 3 3.6 3J9 f2 4.5 i.S 

56 1.4S3 1.492 1.501 1.511 1.520 1.530 1.540 33 10 \ *■» g? 5.6 6.0 6.4 

57 1.540 1.550 1.560 1.570 1.580 1.590 1.600 33 10 % S'g 70 04 „ oq 

58 1.600 1.611 1.621 1.632 1.643 1.653 1.664 3111 7 84 91 98 105 ll".2 

59 1.664 1.675 1.6S6 1.698 1.709 1.720 1.732 30 11 8 9.6 10.4 11.2 12.0 12.8 

9 10.8 11.7 12.6 13.5 14.4 

60 1.732 1.744 1.756 1.767 1.780 1.792 1.804 39 12 

61 1.S04 1.816 1.829 1.842 1.855 1.868 1.881 28 13 
63 1.881 1.894 1.907 1.921 1.935 1.949 1.963 37 14 

63 1.963 1.977 1.991 2.006 2.020 2.035 2.050 36 14 

64 2.050 2.066 2.081 2.097 2.112 2.128 2.145 35 16 

65 2.145 2.161 2.177 2.194 2.211 2.229 2.246 34 17 

66 2.246 2.264 2.282 2.300 2.318 2.337 2.356 33 18 

67 2.356 2.375 2.394 2.414 2.434 2.455 2.475 33 20 
63 2.475 2.496 2.517 2.539 2.560 2.583 2.605 3122 
69 2.605 2.628 2.651 2.675 2.699 2.723 2.747 30 24 





17 


18 


19 


30 


21 


1 


1.7 


1.8 


1.9 


2.0 


2.1 


2 


3.4 


3.6 


3.8 


4.0 


42 


3 


5.1 


5.4 


5.7 


6.0 


6.3 


4 


6.8 


7.2 


7.6 


8.0 


S.4 


5 


8.5 


9.0 


9.5 


100 10.5 


6 10.2 10.8 11.4 12.0 12.6 


7 11.9 12.6 13.3 14.0 14.7 


8 


13.6 14.4 15.2 16.0 16.8 


9 


15.3 16.2 17.1 18.0 18.9 




33 


33 


34 


35 


26 


1 


2.2 


2.3 


2.4 


2.5 


2.6 


2 


4.4 


4.6 


4.8 


5.0 


5.2 


3 


6.6 


6.9 


7.2 


7.5 


7.S 



70 2.747 2.773 2.798 2.824 2.850 2.877 2.90419 26 



9.2 9.6 10.0 10.4 
5 11.0 11.5 12.0 12.5 13.0 



71 2.904 2.932 2.960 2.989 3.018 3.047 3.07818 29 6 132 13 8 144150 156 

72 3.078 3.108 3.140 3.172 3.204 3.237 3.271 17 32 7 15.4 \6.\ \6.% 17.5 18^2 

73 3.271 3.305 3.340 3.376 3.412 3.450 3.487 16 36 8 17.6 18.4 19.2 20.0 20.8 

74 3.487 3.526 3.566 3.606 3.647 3.689 3.732 15 41 9 19-8 20.7 21.6 22.5 23.4 



37 38 59 63 63 

1 2.7 2.8 5.9 6.2 6.3 

2 5.4 5.6 11.8 12.4 12.6 



75 3.732 3.776 3.821 3.867 3.914 3.962 4.01114 46 

76 4.011 4.061 4.113 4.165 4.219 4.275 4.33113 53 

77 4.331 4.390 4.449 4.511 4.574 4.638 4.705 13 62 3 g'j 84 177 18*6 189 

78 4.705 4.773 4.843 4.915 4.989 5.066 5.145 1173 4 10.8 11.2 23^6 24^8 25^ 

79 5.145 5.226 5.309 5.396 5.485 5.576 5.67110 88 5 13.5 14.0 29.5 31.0 31.5 

6 16.2 16.8 35.4 37.2 37.8 

7 18.9 19.6 41.3 43.4 44.1 

80 5.671 5.769 5.871 5.976 6.084 6.197 6.314 9 8 21.6 22.4 47.2 49.6 50.4 

81 6.314 6.435 6.561 6.691 6.827 6.968 7.115 8 9 24.3 25.2 53.155.8 56.7 

82 7.115 7.269 7.429 7.596 7.770 7.953 8.144 7 

83 8.144 8.345 8.556 8.777 9.010 9.255 9.514 6 64 W> 68 70 73 

84 9.514 9.7S8 10.078 10.385 10.712 11.059 11.430 5 1 6.4 6.6 6.8 7.0 7.2 

2 12.8 13.2 13.6 14.0 14.4 

3 19.2 19.8 20.4 21.0 21.6 
8511.43011.826 12.251 12.706 13.197 13.727 14.301 4 4 25.6 26.4 27.2 28.0 28.8 
8614.30114.924 15.605 16.350 17.169 18.075 19.081 3 5 32.0 33.0 34.0 35.0 36.0 

87 19.08120.206 21.470 22.904 24.542 26.432 28.636 3 % ||| **■% !?« 2§n -nl 

88 28.636 31.242 34.368 38.188 42.964 49.104 57.290 1 8 512 52 8 54 4^6 576 

89 57.290 68.750 85.940 114.59 171.89 343.77 infinit. 9 5V6 59^ 61 J e&O %\1 

60' 50' 40' 30' 20' 10' 0' ° d. P. P. 

Numerical values of the cotangent function, 0° to 45°. 



512 



UNIFIED MATHEMATICS 



Radian measure of angles, 0° to 180° 

or 

Length of arc in unit circle for angle 0° to 180° at center. 



A° 


Radians 


A° 


Radians 


A° 


Radians 


A° 


Radians 


1° 

2° 
3° 


0.017 
0.035 
0.052 


46° 

47° 
48° 


0.803 
0.820 
0.838 


91° 
92° 
93° 


1.588 

1.606 
1.623 


136° 
137° 
138° 


2.374 
2.391 
2.409 


4° 
5° 
6° 


0.070 
0.087 
0.105 


49° 
50° 
51° 


0.855 
0.873 
0.890 


94° 
95° 
96° 


1.641 
1.658 

1.676 


139° 
140° 
141° 


2.426 
2.443 
2.461 


7° 
8° 
9° 


0.122 
0.140 
0.157 


52° 
53° 
54° 


0.908 
0.925 
0.942 


97° 
98° 
99° 


1.693 
1.710 
1.728 


142° 
143° 
144° 


2.478 
2.496 
2.513 


10° 
11° 
12° 


0.175 
0.192 
0.209 


55° 
56° 

57° 


0.960 
0.977 
0.995 . 


100° 
101° 
102° 


1.745 
1.763 
1.780 


145° 
146° 
147° 


2.531 

2.548 
2.566 


13° 
14° 
15° 


0.227 
0.244 
0.262 


58° 
59° 
60° 


1.012 
1.031 
1.047 


103° 
104° 
105° 


1.798 
1.815 
1.833 


148° 
149° 
150° 


2.583 
2.601 
2.618 


16° 
17° 

18° 


0.279 
0.297 
0.314 


61° 
62° 
63° 


1.065 
1.082 
1.100 


106° 
107° 
108° 


1.850 

1.868 
1.885 


151° 
152° 
153° 


2.635 
2.653 
2.670 


19° 
20° 
21° 


0.332 
0.349 

0.367 


64° 
65° 
66° 


1.117 
1.134 
1.152 


109° 
110° 
111° 


1.902 
1.920 
1.937* 


154° 
155° 
156° 


2.688 
2.705 
2.723 


22° 
23° 

24° 


0.384 
0.401 
0.419 


67° 
68° 
69° 


1.169 
1.187 
1.204 


112° 
113° 
114° 


1.955 
1.972 
1.990 


157° 
158° 
159° 


2.740 

2.758 
2.775 


25° 
26° 

27° 


0.436 
0.454 
0.471 


70° 
71° 
72° 


1.222 
1.239 
1.257 


115° 
116° 
117° 


2.007 
2.025 
2.042 


160° 

161° 
162° 


2.793 
2.810 

2.827 


28° 
29° 
30° 


0.489 

0.506 
0.524 


73° 

74° 
75° 


1.274 
1.292 
1.309 


118° 
119° 
120° 


2.059 
2.077 
2.094 


163° 
164° 
165° 


2.845 
2.862 
2.880 


31° 
32° 
33° 


0.541 
0.559 
0.576 


76° 

77° 
78° 


1.326 
1.344 
1.361 


121° 
122° 
123° 


2.112 
2.129 
2.147 


166° 
167° 
168° 


2.897 
2.915 
2.932 


34° 
35° 
36° 


0.593 
0.611 
0.628 


79° 
80° 
81° 


1.379 
1.396 
1.414 


124° 
125° 
126° 


2.164 
2.182 
2.199 


169° 
170° 
171° 


2.950 
2.967 
2.985 


37° 
38° 
39° 


0.646 
0.663 
0.681 


82° 
83° 

84° 


1.431 

1.449 
1.466 


127° 
128° 
129° 


2.217 
2.234 
2.251 


172° 
173° 
174° 


3.002 
3.019 
3.037 


40° 
41° 
42° 


0.698 
0.716 
0.733 


85° 
86° 

87° 


1.484 
1.501 
1.518 


130° 
131° 
132° 


2.269 
2.286 
2.304 


175° 
176° 
177° 


3.054 

3.072 
3.089 


43° 
44° 
45° 


0.750 
0.768 
0.785 


88° 
89° 
90° 


1.536 
1.553 
1.571 


133° 
134° 
135° 


2.321 
2.339 
2.356 


178° 
179° 
180° 


3.107 

3.124 
3.142 



TABLES 



513 



Minutes as 


Decimals of 






Growth Function, 


e* 




One Degree or 


Seconds 






Decay Function, e 


— £ 




as 


Decimals 


of One 






e* 


or e l 


e~ x 


jr e~t 


Minute 






X 

t 


log c x 
loge t 


Value 


log,,, 


Value 


log*) 


1 


.017 


31 


.517 


0.0 


00 


1.000 


0.000 


1.000 


0.000 


2 


.033 


32 


.533 


0.1 


-2.303 


1.105 


0.043 


0.905 


9.957 


3 


.050 


33 


.550 


0.2 


-1.610 


1.221 


0.087 


0.819 


9.913 


4 


.067 


34 


.567 


0.3 


-1.204 


1.350 


0.130 


0,741 


9.8T0 


5 


.083 


35 


.583 


0.4 


-0.916 


1.492 


0.174 


0.670 


9.826 


' 6 


.100 


36 


.600 


0.5 


- 0.693 


1.649 


0.217 


0,607 


0.783 


7 


.117 


37 


.617 


0.6 


-0.511 


1.822 


0.261 


0.549 


9.739 


8 


.133 


38 


.633 


0.7 


-0.357 


2.014 


0.304 


497 


9.696 


9 


.150 


39 


.650 


0.8 


-0.223 


2.226 


0.347 


0.449 


9.653 


10 


.167 


40 


.667 


0.9 


-0.105 


2.460 


0.391 


0.407 


9.609 


11 


.183 


41 


.683 


1.0 


0.000 


2.718 


0.434 


0.368 


9.566 


12 


.200 


42 


.700 


1.1 


0.095 


3.004 


0.478 


0.333 


9.522 


13 


.217 


43 


.717 


1.2 


0.182 


3.320 


0.521 


0.301 


9.479 


14 


.233 


44 


.733 


1.3 


0.262 


3.769 


0.565 


0.273 


9.435 


15 


.250 


45 


.750 


1.4 


0.336 


4.055 


0.608 


0.247 


9.392 


16 


.267 


46 


.767 


1.5 


0.405 


4.482 


0.651 


0.223 


9.349 


17 


.283 


47 


.783 


1.6 


0.470 


4.953 


0.695 


0.202 


9.305 


18 


.300 


48 


.800 


1.7 


0.531 


5.474 


0.738 


0.183 


9.262 


19 


.317 


49 


.817 


1.8 


0.588 


6.050 


0.782 


0.165 


9.218 


20 


.333 


50 


.833 


1.9 


0.642 


6.686 


0.825 


0.150 


9.175 


21 


.350 


51 


.850 


2.0 


0.693 


7.3S9 


0.869 


0.135 


9.131 


22 


.367 


52 


.867 


2.1 


0.742 


8.166 


0.912 


0.122 


9.088 


23 


.383 


53 


.883 


2.2 


0.788 


9.025 


0.955 


0.111 


9.045 


24 


.400 


54 


.900 


2.3 


0.833 


9.974 


0.999 


0.100 


0.001 


25 


.417 


55 


.917 


2.4 


0.S75 


11.02 


1.023 


0.091 


8.958 


26 


.433 


56 


.933 


2.5 


0.916 


12.18 


1.086 


0.082 


8.914 


27 


.450 


57 


.950 


2.6 


0.956 


13.46 


1.129 


0.074 


8.871 


28 


.467 


58 


.967 


2.7 


0.993 


14.88 


1.173 


0.067 


8.827 


29 


.483 


59 


.983 


2.8 


1.030 


16.44 


1.216 


0.061 


8.784 


30 


.500 


60 


1.000 


2.9 


1.065 


18.17 


1.259 


0.055 


8.741 










3.0 


1.099 


20.09 


1.303 


0.050 


8.697 










3.1 


1.132 


22.20 


1.346 


0.045 


8.654 










3.2 


1.163 


24.53 


1.390 


0.041 


8.610 










3.3 


1.193 


27.11 


1.433 


0.037 


8.567 










3.4 


1.224 


29.96 


1.477 


0.033 


8.523 










3.5 


1.253 


33.12 


1.520 


0.030 


8.480 










4.0 


1.386 


54.60 


1.737 


0.018 


8.263 










4.5 


1.504 


90.02 


1.954 


0.0111 


8.046 










5.0 


1.609 


148.4 


2.171 


0.0067 


7.829 










6.0 


1.792 


408.4 


2.606 


0.0025 


7.394 










7.0 


1.946 


1096.6 


3.040 


0.0009 


6.960 










8.0 


2.079 


2981.0 


3.474 


0.0003 


6.526 










9.0 


2.197 


8103.1 


3.909 


0.0001 


6.091 










10.0 


2.303 


22026. 


4.343 


0.0000 


5.657 



514 UNIFIED MATHEMATICS 

The accumulation of 1 at the end of n years, r* 1 = (1 + £)». 



Years. 11%. 



2%. 



1 


1.0150 


1.0200 


2 


1.0302 


1.0404 


3 


1.0457 


1.0612 


4 


1.0614 


1.0824 


5 


1.0773 


1.1041 


6 


1.0934 


1.1262 


7 


1.1098 


1.1487 


8 


1.1265 


1.1717 


9 


1.1434 


1.1951 


10 


1.1605 


1.2190 


11 


1.1779 


1.2434 


13 


1.1956 


1.2682 


13 


1.2136 


1.2936 


14 


1.2318 


1.3195 


15 


1.2502 


1.3459 


16 


1.2690 


1.3728 


17 


1.2880 


1.4002 


18 


1.3073 


1.4282 


19 


1.3270 


1.4568 


20 


1.3469 


1.4859 


21 


1.3671 


1.5157 


22 


1.3876 


1.5460 


23 


1.4084 


1.5769 


24 


1.4295 


1.6084 


25 


1.4509 


1.6406 


26 


1.4727 


1.6734 


27 


1.4948 


1.7069 


28 


1.5172 


1.7410 


29 


1.5400 


1.7758 


30 


1.5631 


1.8114 


31 


1.5865 


1.8476 


32 


1.6103 


1.8845 


33 


1.6345 


1.9222 


34 


1.6590 


1.9607 


35 


1.6839 


1.9999 


36 


1.7091 


2.0399 


37 


1.7348 


2.0807 


38 


1.7608 


2.1223 


39 


1.7872 


2.1647 


40 


1.8140 


2.2080 


50 


2.1052 


2.6916 


60 


2.4432 


3.2810 


70 


2.8355 


3.9996 


80 


3.2907 


4.8754 


90 


3.8190 


5.9431 


100 


4.4321 


7.2447 



Years. 11%. 2%, 



21%. 

1.0250 
1.0506 
1.0769 
1.1038 
1.1314 



1.1597 
1.1887 
1.2184 
1.2489 
1.2801 

1.3121 
1.3449 
1.3785 
1.4130 
1.4483 

1.4845 
1.5216 
1.5597 
1.5987 
1.6386 

1.6796 
1.7216 
1.7646 
1.8087 
1.8539 

1.9003 
1.9478 
1.9965 
2.0464 
2.0976 

2.1500 
2.2038 
2.2589 
2.3153 
2.3732 

2.4325 
2.4933 
2.5557 
2.6196 
2.6851 

3.4371 
4.3998 
5.6321 
7.2096 
9.2289 
11.8137 

21%. 



3%. 

1.0300 
1.0609 
1.0927 
1.1255 
1.1593 

1.1941 
1.2299 
1.2668 
1.3048 
1.3439 

1.3842 
1.4258 
1.4685 
1.5126 
1.5580 

1.6047 
1.6528 
1.7024 
1.7535 
1.8061 

1.8603 
1.9161 
1.9736 
2.0328 
2.0938 

2.1566 
2.2213 
2.2879 
2.3566 
2.4273 

2.5001 
2.5751 
2.6523 
2.7319 
2.8139 

2.8983 
2.9852 
3.0748 
3.1670 
3.2620 

4.3839 

5.8916 

7.9178 

10.6409 

14.3005 

19.2186 

3%. 



4%. 


5%. 


6%. Years 


1.0400 


1.0500 


1.0600 


1 


1.0816 


1.1025 


1.1236 


2 


1.1249 


1.1576 


1.1910 


3 


1.1699 


1.2155 


1.2625 


4 


1.2167 


1.2763 


1.3382 


5 


1.2653 


1.3401 


1.4185 


6 


1.3159 


1.4071 


1.5036 


7 


1.3686 


1.4775 


1.5938 


8 


1.4233 


1.5513 


1.6895 


9 


1.4802 


1.6289 


1.7908 


10 


1.5395 


1.7103 


1.8983 


11 


1.6010 


1.7959 


2.0122 


12 


1.6651 


1.8856 


2.1329 


13 


1.7317 


1.9799 


2.2609 


14 


1.8009 


2.0789 


2.3966 


15 


1.8730 


2.1829 


2.5404 


16 


1.9473 


2.2920 


2.6928 


17 


2.0258 


2.4066 


2.8543 


18 


2.1068 


2.5270 


3.0256 


19 


2.1911 


2.6533 


3.2071 


20 


2.2788 


2.7860 


3.3996 


21 


2.3699 


2.9253 


3.6035 


22 


2.4647 


3.0715 


3.8197 


23 


2.5633 


3.2251 


4.0489 


24 


2.6658 


3.3864 


4.2919 


25 


2.7725 


3.5557 


4.5494 


26 


2.8834 


3.7335 


4.8223 


27 


2.9987 


3.9201 


5.1117 


28 


3.1187 


4.1161 


5.4184 


29 


3.2434 


4.3219 


5.7435 


30 


3.3731 


4.5380 


6.0881 


31 


3.5081 


4.7649 


6.4534 


32 


3.6484 


5.0032 


6.8406 


33 


3.7943 


5.2533 


7.2510 


34 


3.9461 


5.5160 


7.6861 


35 


4.1039 


5.7918 


8.1473 


36 


4.2681 


6.0814 


8.6361 


37 


4.4388 


6.3855 


9.1543 


38 


4.6164 


6.7048 


9.7035 


39 


4.8010 


7.0400 


10.2857 


40 


7.1067 


11.4674 


18.4202 


50 


10.5196 


18.6792 


32.9877 


60 


15.5716 


30.4264 


59.0759 


70 


23.0498 


49.6514 


105.7960 


80 


34.1193 


80.7304 


189.4645 


90 


50.5050 


131.5013 


339.3021 


100 


4%. 


5%. 


6%. 


Sfears 



TABLES 



515 



The present value of 1 due in n years. v n = (1 + iy 



Years. \\%. 



2%. 



2h%- 



1 


0.9852 


0.9804 


0.9756 


2 


0.9707 


0.9612 


0.9518 


3 


0.9563 


0.9423 


0.9286 


4 


0.9422 


0.923S 


0.9060 


5 


0.9283 


0.9057 


0.8839 


6 


0.9145 


0.8880 


0.8623 


7 


0.9010 


0.S706 


0.8413 


8 


0.8877 


0.8535 


0.8207 


9 


0.8746 


0.8368 


0.8007 


10 


0.8617 


0.8203 


0.7812 


11 


0.8489 


0.8043 


0.7621 


12 


0.8364 


0.7885 


0.7436 


13 


0.8240 


0.7730 


0.7254 


14 


0.8118 


0.7579 


0.7077 


15 


0.7999 


0.7430 


0.6905 


16 


0.7880 


0.7284 


0.6736 


17 


0.7764 


0.7142 


0.6572 


18 


0.7649 


0.7002 


0.6412 


19 


0.7536 


0.6864 


0.6255 


20 


0.7425 


0.6730 


0.6103 


21 


0.7315 


0.6598 


0.5954 


22 


0.7207 


0.6468 


0.5809 


23 


0.7100 


0.6342 


0.5667 


24 


0.6995 


0.6217 


0.5529 


25 


0.6892 


0.6095 


0.5394 


26 


0.6790 


0.5976 


0.5262 


27 


0.6690 


0.5859 


0.5134 


28 


0.6591 


0.5744 


0.5009 


29 


0.6494 


0.5631 


0.4887 


30 


0.6398 


0.5521 


0.4767 


31 


0.6303 


0.5412 


0.4651 


32 


0.6210 


0.5306 


0.4538 


33 


0.6118 


0.5202 


0.4427 


34 


0.6028 


0.5100 


0.4319 


35 


0.5939 


0.5000 


0.4214 


36 


0.5851 


0.4902 


0.4111 


37 


0.5764 


0.4806 


0.4011 


38 


0.5679 


0.4712 


0.3913 


39 


0.5595 


0.4620 


0.3817 


40 


0.5513 


0.4529 


0.3724 


50 


0.4750 


0.3715 


0.2909 


60 


0.4093 


0.3048 


0.2273 


70 


0.3527 


0.2500 


0.1776 


80 


0.3039 


0.2051 


0.1387 


90 


0.2619 


0.1683 


0.1084 


100 


0.2256 


0.1380 


0.0846 



Years. \\% 



2%. 



21%. 



3%. 

0.9709 
0.9426 
0.9151 
0.8885 
0.8626 

0.8375 
0.8131 
0.7894 
0.7664 
0.7441 

0.7224 
0.7014 
0.6810 
0.6611 
0.6419 

0.6232 
0.6050 
0.5874 
0.5703 
0.5537 

0.5375 
0.5219 
0.5067 
0.4919 
0.4776 

0.4637 
0.4502 
0.4371 
0.4243 
0.4120 

0.4000 
0.3883 
0.3770 
0.3660 
0.3554 

0.3450 
0.3350 
0.3252 
0.3158 
0.3066 

0.2281 
0.1697 
0.1263 
0.0940 
0.0699 
0.0520 

3%. 



4%. 



5%. 



6%. 



Years. 



0.9615 


0.9524 


0.9434 


1 


0.9246 


0.9070 


0.8900 


2 


0.8890 


0.8638 


0.8396 


3 


0.8548 


0.8227 


0.7921 


4 


0.8219 


0.7835 


0.7473 


5 


0.7903 


0.7462 


0.7050 


6 


0.7599 


0.7107 


0.6651 


7 


0.7307 


0.6768 


0.6274 


8 


0.7026 


0.6446 


0.5919 


9 


0.6756 


0.6139 


0.5584 


10 


0.6496 


0.5847 


0.5268 


11 


0.6246 


0.5568 


0.4970 


12 


0.6006 


0.5303 


0.4688 


13 


0.5775 


0.5051 


0.4423 


14 


0.5553 


0.4810 


0.4173 


15 


0.5339 


0.4581 


0.3936 


16 


0.5134 


0.4363 


0.3714 


17 


0.4936 


0.4155 


0.3503 


18 


0.4746 


0.3957 


0.3305 


19 


0.4564 


0.3769 


0.3118 


20 


0.4388 


0.3589 


0.2942 


21 


0.4220 


0.3418 


0.2775 


22 


0.4057 


0.3256 


0.2618 


23 


0.3901 


0.3101 


0.2470 


24 


0.3751 


0.2953 


0.2330 


25 


0.3607 


0.2812 


0.2198 


26 


0.3468 


0.2678 


0.2074 


27 


0.3335 


0.2551 


0.1956 


28 


0.3207 


0.2429 


0.1846 


29 


0.3083 


0.2314 


0.1741 


30 


0.2965 


0.2204 


0.1643 


31 


0.2851 


0.2099 


0.1550 


32 


0.2741 


0.1999 


0.1462 


33 


0.2636 


0.1904 


0.1379 


34 


0.2534 


0.1813 


0.1301 


35 


0.2437 


0.1727 


0.1227 


36 


0.2343 


0.1644 


0.1158 


37 


0.2253 


0.1566 


0.1092 


38 


0.2166 


0.1491 


0.1031 


39 


0.2083 


0.1420 


0.0972 


40 


0.1407 


0.0872 


0.0543 


50 


0.0951 


0.0535 


0.0303 


60 


0.0642 


0.0329 


0.4)169 


70 


0.0434 


0.0202 


0.0095 


80 


0.0293 


0.0124 


0.0053 


90 


0.0198 


0.0076 


0.0029 


100 



4%. 



5%. 



6%. 



Years. 



516 



UNIFIED MATHEMATICS 



The accumulation of an annuity of 1 per annum at the end of n years. 

_ (1 + *)" - 1 
S 7T)- 1 



Years. 1J%. 



1.0000 
2.0150 
3.0452 
4.0909 
5.1523 



6 6.2296 

7 7.3230 

8 8.4328 

9 9.5593 

10 10.7027 

11 11.8633 

12 13.0412 

13 14.2368 

14 15.4504 

15 16.6821 

16 17.9324 

17 19.2014 

18 20.4894 

19 21.7967 
23.1237 



20 



2%. 

1.0000 
2.0200 
3.0604 
4.1216 
5.2040 



6.3081 
7.4343 
8.5830 
9.7546 
10.9497 

12.1687 
13.4121 
14.6803 
15.9739 
17.2934 

18.6393 
20.0121 
21.4123 
22.8406 
24.2974 



21%. 

1.0000 
2.0250 
3.0756 
4.1525 
5.2563 

6.3877 
7.5474 
8.7361 
9.9545 
11.2034 

12.4835 
13.7956 
15.1404 
16.5190 
17.9319 

19.3802 
20.8647 
22.3863 
23.9460 
25.5447 



3%. 

1.0000 
2.0300 
3.0909 
4.1836 
5.3091 

6.4684 

7.6625 

8.8923 

10.1591 

11.4638 

12.8078 
14.1920 
15.6178 
17.0863 
18.5989 

20.1569 
21.7616 
23.4144 
25.1169 
26.8704 



4%. 

1.0000 
2.0400 
3.1216 
4.2465 
5.4163 

6.6330 

7.8983 

9.2142 

10.5828 

12.0061 

13.4864 
15.0258 
16.6268 
18.2919 
20.0236 

21.8245 
23.6975 
25.6454 
27.6712 
29.7781 



5%. 

1.0000 
2.0500 
3.1525 
4.3101 
5.5256 

6.8019 

8.1420 

9.5491 

11.0266 

12.5779 

14.2068 
15.9171 
17.7130 
19.5986 
21.5786 

23.6575 
25.8404 
28.1324 
30.5390 
33.0660 



6%. Years. 



1.0000 
2.0600 
3.1836 
4.3746 
5.6371 

6.9753 

8.3938 

9.8975 

11.4913 

13.1808 

14.9716 
16.8699 
18.8821 
21.0151 
23.2760 



1 
2 
3 
4 
5 

6 
7 

8 

9 

10 

11 
12 
13 
14 
15 

25.6725 16 
28.2129 17 
30.9057 18 
33.7600 
36.7856 



19 
20 



21 24.4705 

22 25.8376 

23 27.2251 

24 28.6335 

25 30.0630 



25.7833 
27.2990 
28.8450 
30.4219 
32.0303 



27.1833 
28.8629 
30.5844 
32.3490 
34.1578 



28.6765 
30.5368 
32.4529 
34.4265 
36.4593 



31.9692 
34.2480 
36.6179 
39.0826 
41.6459 



35.7193 
38.5052 
41.4305 
44.5020 
47.7271 



39.9927 
43.3923 
46.9958 
50.8156 
54.8645 



21 
22 
23 
24 
25 



26 31.5140 

27 32.9867 

28 34.4815 

29 35.9987 

30 37.5387 

31 39.1018 

32 40.6883 

33 42.2986 

34 43.9331 

35 45.5921 



33.6709 
35.3443 
37.0512 
38.7922 
40.5681 

42.3794 
44.2270 
46.1116 
48.0338 
49.9945 



36.0117 
37.9120 
39.8598 
41.8563 
43.9027 

46.0003 
48.1503 
50.3540 
52.6129 
54.9282 



36 47.2760 51.9944 57.3014 

37 48.9851 54.0343 59.7339 

38 50.7199 56.1149 62.2273 

39 52.4807 58.2372 64.7830 

40 54.2679 60.4020 67.4026 

50 73.6828 84.5794 97.4843 

60 96.2147 114.0515 135.9916 

70 122.3638 149.9779 185.2841 

152.7109 193.7720 248.3827 

187.9299 247.1567 329.1543 



80 
90 

100 228.8030 312.2323 432.5487 



38.5530 
40.7096 
42.9309 
45.2189 
47.5754 

50.0027 
52.5028 
55.0778 
57.7302 
60.4620 

63.2759 
66.1742 
69.1594 
72.2342 
75.4013 

112.7969 
163.0534 
230.5941 
321.3630 
443.3489 
607.2877 



Years. 1\%. 



2%. 



%h%. 3% 



44.3117 
47.0842 
49.9676 
52.9663 
56.0849 

59.3283 
62.7015 
66.2095 
69.8579 
73.6522 

77.5983 
81.7022 
85.9703 
90.4092 
95.0255 

152.6671 
237.9907 
364.2905 
551.2450 
827.9833 
1237.6237 

4%. 



51.1135 
54.6691 
58.4026 
62.3227 
66.4389 



59.1564 
63.7058 
68.5281 
73.6398 
79.0582 



95.8363 
101.6281 
107.7095 
114.0950 
120.7998 

209.3480 
353.5837 
588.5285 
971.2288 
1594.6073 
2610.0252 

5%. 



119.1209 
127.2681 
135.9042 
145.0585 
154.7620 

290.3359 

533.1282 

967.9322 

1746.5999 

3141.0752 

5638.3681 



26 
27 

28 
29 
30 



70.7608 84.8017 31 

75.2988 90.8898 32 

80.0638 97.3432 33 

85.0670 104.1838 34 

90.3203 111.4348 35 



37 

38 
39 
40 

50 
60 
70 
80 
90 
100 



6%. Years. 



TABLES 



517 



The present value of an annuity of 1 for n years, 



» 



Years. 1| ( 



2%. 



21%. 



3%. 



4%. 



5%. 



6%. Years. 



1 


0.9S52 


0.9804 


0.9756 


0.9709 


0.9615 


0.9524 


0.9434 


1 


2 


1.9559 


1.9416 


1.9274 


1.9135 


1.8861 


1.8594 


1.8334 


2 


3 


2.9122 


2.SS39 


2.S560 


2.8286 


2.7751 


2.7232 


2.6730 


3 


4 


3.S544 


3.S077 


3.7620 


3.7171 


3.6299 


3.5460 


3.4651 


4 


5 


4.7827 


4.7135 


4.6458 


4.5797 


4.451S 


4.3295 


4.2124 


5 


6 


5.6972 


5.6014 


5.5081 


5.4172 


5.2421 


5.0757 


4.9173 


6 


7 


6.5982 


6.4720 


6.3494 


6.2303 


6.0021 


5.7864 


5.5824 


7 


8 


7.4S59 


7.3255 


7.1701 


7.0197 


6.7327 


6.4632 


6.2098 


8 


9 


S.3605 


8.1622 


7.9709 


7.7861 


7.4353 


7.107S 


6.8017 


9 


10 


9.2222 


8.9826 


8.7521 


8.5302 


8.1109 


7.7217 


7.3601 


10 


11 


10.0711 


9.786S 


9.5142 


9.2526 


8.7605 


8.3064 


7.8869 


11 


12 


10.9075 


10.5753 


10.2578 


9.9540 


9.3851 


8.8633 


8.3838 


12 


13 


11.7315 


11.3484 


10.9832 


10.6350 


9.9856 


9.3936 


8.8527 


13 


14 


12.5434 


12.1062 


11.6909 


11.2961 


10.5631 


9.8986 


9.2950 


14 


15 


13.3432 


12.8493 


12.3814 


11.9379 


11.1184 


10.3797 


9.7122 


15 


16 


14.1313 


13.5777 


13.0550 


12.5611 


11.6523 


10.8378 


10.1059 


16 


17 


14.9076 


14.2919 


13.7122 


13.1661 


12.1657 


11.2741 


10.4773 


17 


18 


15.6726 


14.9920 


14.3534 


13.7535 


12.6593 


11.6896 


10.8276 


18 


19 


16.4262 


15.6785 


14.9789 


14.3238 


13.1340 


12.0853 


11.1581 


19 


20 


17.1686 


16.3514 


15.5892 


14.8775 


13.5903 


12.4622 


11.4699 


20 


21 


17.9001 


17.0112 


16.1845 


15.4150 


14.0292 


12.8212 


11.7641 


21 


22 


18.6208 


17.6580 


16.7654 


15.9369 


14.4511 


13.1630 


12.0416 


22 


23 


19.3309 


18.2922 


17.3321 


16.4436 


14.8568 


13.4886 


12.3034 


23 


11 


20.0304 


18.9139 


17.8850 


16.9355 


15.2470 


13.7986 


12.5504 


24 


25 


20.7196 


19.5235 


18.4244 


17.4131 


15.6221 


14.0940 


12.7834 


25 


26 


21.3986 


20.1210 


18.9506 


17.8768 


15.9828 


14.3752 


13.0032 


26 


27 


22.0676 


20.7069 


19.4640 


18.3270 


16.3296 


14.6430 


13.2105 


27 


28 


22.7267 


21.2813 


19.9649 


18.7641 


16.6631 


14.8981 


13.4062 


28 


29 


23.3761 


21.S444 


20.4535 


19.1885 


16.9837 


15.1411 


13.5907 


29 


30 


24.0158 


22.3965 


20.9303 


19.6004 


17.2920 


15.3725 


13.7648 


30 


31 


24.6461 


22.9377 


21.3954 


20.0004 


17.5885 


15.5928 


13.9291 


31 


32 


25.2671 


23.4683 


21.8492 


20.3888 


17.8736 


15.8027 


14.0840 


32 


33 


25.8790 


23.9886 


22.2919 


20.765S 


18.1476 


16.0025 


14.2302 


33 


34 


26.4817 


24.4986 


22.7238 


21.1318 


18.4112 


16.1929 


14.3681 


34 


35 


27.0756 


24.9986 


23.1452 


21.4872 


18.6646 


16.3742 


14.4982 


35 


36 


27.6607 


25.4888 


23.5563 


21.8323 


18.9083 


16.5469 


14.6210 


36 


37 


28.2371 


25.9695 


23.9573 


22.1672 


19.1426 


16.7113 


14.7368 


37 


38 


28.8051 


26.4406 


24.3486 


22.4925 


19.3679 


16.8679 


14.8460 


38 


39 


29.3646 


26.9026 


24.7303 


22.8082 


19.5845 


17.0170 


14.9491 


39 


40 


29.9158 


27.3555 


25.1028 


23.1148 


19.7928 


17.1591 


15.0463 


40 


50 


34.9997 


31.4236 


28.3623 


25.7298 


21.4822 


18.2559 


15.7619 


50 


60 


39.3803 


34.7609 


30.9087 


27.6756 


22.6235 


18.9293 


16.1614 


60 


70 


43.1.549 


37.4987 


32.8979 


29.1234 


23.3945 


19.3427 


16.3845 


70 


80 


46.4073 


39.7445 


34.4518 


30.2008 


23.9154 


19.5965 


16.5091 


80 


90 


49.2099 


41.5869 


35.6658 


31.0024 


24.2673 


19.7523 


16.5787 


90 


100 


51.6247 


43.0983 


36.6141 


31.5989 


24.5050 


19.8479 


16.6175 


100 



Years. li%. 



2%. 






3%. 



4 



5%, 



6%. Years. 



518 



UNIFIED MATHEMATICS 



The annual sinking fund which will accumulate to 1 at the end of n years. 
% To obtain — add i, since — = — 



tei (i + i) w -i 



+ 1. 



Years. \\%. 



1.0000 
0.4963 
0.3284 
0.2444 
0.1941 



2%. 

1.0000 
0.4950 
0.3268 
0.2426 
0.1922 



1.0000 
0.4938 
0.3251 
0.2408 
0.1902 



3%. 

1.0000 
0.4926 
0.3235 
0.2390 
0.1884 



4%. 

1.0000 
0.4902 
0.3203 
0.2355 
0.1846 



5%. 

1.0000 
0.4878 
0.3172 
0.2320 
0.1810 



6%. Years. 



1.0000 
0.4854 
0.3141 
0.2286 
0.1774 



0.1605 
0.1366 
0.1186 
0.1046 
0.0934 



0.1585 
0.1345 
0.1165 
0.1025 
0.0913 



0.1566 
0.1325 
0.1145 
0.1005 
0.0893 



0.1546 
0.1305 
0.1125 
0.0984 
0.0872 



0.1508 
0.1266 
0.1085 
0.0945 
0.0833 



0.1470 
0.1228 
0.1047 
0.0907 
0.0795 



0.1434 
0.1191 
0.1010 
0.0870 
0.0759 



0.0843 
0.0767 
0.0702 
0.0647 
0.0599 



0.0822 
0.0746 
0.0681 
0.0626 
0.0578 



0.0801 
0.0725 
0.0660 
0.0605 
0.0558 



0.0781 
0.0705 
0.0640 
0.0585 
0.0538 



0.0741 
0.0666 
0.0601 
0.0547 
0.0499 



0.0704 
0.0628 
0.0565 
0.0510 
0.0463 



0.0668 
0.0593 
0.0530 
0.0476 
0.0430 



16 
17 

18 
19 
20 

21 
22 
23 
24 



0.0558 
0.0521 
0.0488 
0.0459 
0.0432 

0.0409 
0.0387 
0.0367 
0.0349 
0.0333 



0.0537 
0.0500 
0.0467 
0.0438 
0.0412 

0.0388 
0.0366 
0.0347 
0.0329 
0.0312 



0.0516 
0.0479 
0.0447 
0.0418 
0.0391 

0.0368 
0.0346 
0.0327 
0.0309 
0.0293 



0.0496 
0.0460 
0.0427 
0.0398 
0.0372 

0.0349 
0.0327 
0.0308 
0.0290 
0.0274 



0.0458 
0.0422 
0.0390 
0.0361 
0.0336 

0.0313 
0.0292 
0.0273 
0.0256 
0.0240 



0.0423 
0.0387 
0.0355 
0.0327 
0.0302 

0.0280 
0.0260 
0.0241 
0.0225 
0.0210 



0.0390 
0.0354 
0.0324 
0.0296 
0.0272 

0.0250 
0.0230 
0.0213 
0.0197 
0.0182 



20 

27 
28 
29 
30 

31 
32 
33 
34 
35 



0.0317 
0.0303 
0.0290 
0.0278 
0.0266 

0.0256 
0.0246 
0.0236 
0.0228 
0.0219 



0.0297 
0.0283 
0.0270 
0.0258 
0.0246 

0.0236 
0.0226 
0.0217 
0.0208 
0.0200 



0.0278 
0.0264 
0.0251 
0.0239 
0.0228 

0.0217 
0.0208 
0.0199 
0.0190 
0.0182 



0.0259 
0.0246 
0.0233 
0.0221 
0.0210 

0.0200 
0.0190 
0.0182 
0.0173 
0.0165 



0.0226 
0.0212 
0.0200 
0.0189 
0.0178 

0.0169 
0.0159 
0.0151 
0.0143 
0.0136 



0.0196 
0.0183 
0.0171 
0.0160 
0.0151 

0.0141 
0.0133 
0.0125 
0.0118 
0.0111 



0.0169 
0.0157 
0.0146 
0.0136 
0.0126 

0.0118 
0.0110 
0.0103 
0.0096 
0.0090 



29 



37 



0.0212 
0.0204 
0.0197 
0.0191 
0.0184 



0.0192 
0.0185 
0.0178 
0.0172 
0.0166 



0.0175 
0.0167 
0.0161 
0.0154 
0.0148 



0.0158 
0.0151 
0.0145 
0.0138 
0.0133 



0.0129 
0.0122 
0.0116 
0.0111 
0.0105 



0.0104 
0.0098 
0.0093 
0.0088 
0.0083 



0.0084 
0.0079 
0.0074 
0.0069 
0.0065 



50 
60 
70 
80 
90 
109 



0.0136 
0.0104 
0.0182 
0.0065 
0.0053 
0.0044 



Years. \\%. 



0.0118 
0.0088 
0.0067 
0.0052 
0.0040 
0.0032 

3%. 



0.0103 
0.0074 
0.0054 
0.0040 
0.0030 
0.0023 

2*%. 



0.0089 
0.0061 
0.0043 
0.0031 
0.0023 
0.0016 

3%. 



0.0066 

0.0042 

0.0027 

0.0018 

0.00121 

0.00081 

4%. 



0.0048 

0.0028 

0.0017 

0.0010 

0.00063 

0.00038 

5%. 



0.0034 

0.0019 

0.0010 

0.0006 

0.00032 

0.00018 



50 
60 
70 
80 
90 
100 



6%. Years. 



INDEX 



Abel, 204, 399. 
abscissa, 67. 
absolute value, 104. 
addition, formulas, 237. 

geometrical, 13. 

of numbers, 4. 
aeroplane, 277, 278, 388. 
Ahmes papyrus, 90. 
air-pump, 179 ff. 
Al-Battani, 126, 216. 
Alexander III Bridge in Paris, 309. 
algebra, fundamental theorem, 392. 

literal, 12. 
algebraic functions, 66 ff., 392 ff. 
Almagest, 125. 
amplitude, of complex number, 444. 

of S. H. M., 412. 

of sinusoid, 408, 410. 
angle, between two lines, 246 ff., 461. 

depression, 210. 

direction, 457. 

measurement, 110 ff. 

of incidence, 262 ff. 

of refraction, 262 ff. 
annuity, 183 ff. 

approximations, logarithms, 48, 140 
ff., 274. 

numerical, 31 ff., 93. 

trigonometric, 170 ff. 
anti-sine, -cosine, -tangent, etc., 137. 
Arabic mathematics, 90, 119, 126, 

203, 256. 
Archimedes, 183, 404, 438. 
arc sin, -cos, -tan, etc., 137. 
area, of an ellipse, 287 ff. 

of an inscribed quadrilateral, 261. 

of a triangle, 51, 209 ff., 258 ff., 261. 
Argand, 450. 
arithmetical series, 166 ff. 

mean, 172 ff. 
asymptote, 325. 
auxiliary circles, 282 ff. 



Babylonian mathematics, 111, 183. 
bacteria, growth, 423, 427. 
barometric pressure, 60, 426 ff. 
Bhaskara, 261. 
binomial series, 193 ff. 
biquadratic, 392, 399. 
bissctor, of the angle between two 
lines, 160 ff. 
perpendicular, 151, 164. 
Briggs, 49. 

capacity, of a can, 53, 73, 83, 100. 

of a cistern, 94. 
Cardan, 399. 
cardioid, 437. 
Carrel, 430. 

centigrade scale, 2, 86, 103. 
centimeters and inches, 103. 
characteristic, 44. 
chess-board problem, 187. 
Chinese mathematics, 203. 
circle, 220 ff. 

auxiliary, 282 ff. 
circular sections of a cone, etc., 489 ff. 
circumferential velocity, 115. 
cissoid, 437. 

Colosseum in Rome, 288, 362. 
compass, geometrical, 4. 

mariner's, 114. 
complex numbers, 439 ff. 
components of a vector, 152 ff. 
compound interest, 54 ff . 
conchoid of Nicomedes, 437. 
congruent angles, 116. 
conjugate hyperbolas, 328. 

numbers, 441. 
cone, 483. 

conic sections, 287, 483 ff. 
connecting rod, 421 ff. 
continuous functions, 393, 397. 
coordinate axes, 453. 

planes, 453. 



519 



520 



INDEX 



coordinates, 67 ff., 115 ff., 118, 435, 

453. 
cosecant, definition, 120. 
cosine, definition, 117 ff. 

law, 252 ff . 
cotangent, definition, 120, 125 ff. 
crank arm, 421 ff. 
cubic, 392, 399. 

curves, 466. 
cubical parabola, 74. 
curvature of the earth, 146. 
cyclic interchange, 252. 
cylindrical surfaces, 466 ff. 

damped vibrations, 431 ff. 
decagon, 124. 
decimal, recurring, 177. 
deflection angle, 217. 
De Moivre, 445. 
departure, 210. 
depression, 210. 
Descartes, 69, 449. 
dihedral angle, 456. 
dip, 210. 
directed line, 1. 
direction angles, 457, 470. 

cosines, 457, 462. 
directrix, of a conic, 289, 309, 320. 

of a cylinder, 467. 
discount, 190, 386 ff. 
discriminant, 89. 

distance, between two points, 104 ff., 
458. 

from a point to a line, 158 ff., 487. 

from a point to a plane, 471. 
division, 5. 

abbreviated, 35. 

graphical, 9, 10. 

synthetic, 25, 394 ff. 
duplication of a cube, 204 ff . 

eccentricity (e), 289, 297, 347. 
Egyptian mathematics, 90, 175, 183. 
electrical phenomena, 108, 412, 

415 ff., 432. 
element of a cylinder, 467. 
ellipse, 280 ff. 
ellipsoids, 476 ff. 
elliptical arch, 295, 307, 355. 

gears, 356 ff. 
elliptic paraboloid, 481 ff. 
equilateral hyperbola, 327, 329. 
Euclid, 183. 



Euler, 243. 
evolution, 41. 
explicit function, 58. 
exponent, 40. 

Fahrenheit scale, 2, 86, 103. 
family of surfaces, 465 ff . 
Farm Loan Act, 188 ff. 
Fermat, 69. 
Ferrari, 399. 
Fiori, 399. 
fly-wheel, 434. 
focal distances, 291, 327. 
focus, 289, 299 ff., 305, 309. 
fourth dimension, 452 ff. 
Franklin, 56. 
frequency, 412. 
functions, 58. 

algebraic, 66 ff. 

linear, 77 ff., 101 ff., 149 ff., 155 ff., 
462 ff. 

quadratic, 87 ff. 

trigonometric, 110 ff. 

Galton, 179. 

Gauss, 392, 450. 

generator, of a cylinder, 467. 

of a surface^ 491. 
geometrical mean, 178. 

series, 176 ff. 
Glover, 188. 

graphical methods, 13 ff., 169ff.,180ff. 
greater, 1, 3. 

Greek mathematics, 119, 125, 183, 
242, 256 ff., 287. 

half-angle formulas, trigonometric 
functions, 247 ff. 

oblique triangle, 258. 
Halley, 428. 

healing of a wound, law, 429 ff. 
helical spring, 414 ff. 
Hero, 51, 69, 261. 
Hill Auditorium, 352, 362, 489. 
Hindu mathematics, 119, 261. 
hyperbola, 320 ff. 
hyperbolic paraboloid, 481 ff. 
hyperboloid, of one sheet, 475, 479 ff. 

of two sheets, 475, 479 ff. 

imaginary numbers, 439 ff., 449, 450. 
implicit function, 58. 
index of refraction, 262 ff. 
indices, theory of, 40 ff. 



INDEX 



521 



induction, mathematical, 167, 198. 
infinity, 97 ff., 176 ff., 181, 200 ff. 

311 ff., 323 ff. 
integers, 3. 
intercepts. 82 ff. 

interest, 54 ff., 92, 183 ff., 200, 423. 
interpolation, 46 ff.. 140 ff., 170 ff. 
intersections of graphs, SO ff., 465. 
inverse functions, 137. 
involution, 41. 
irrational numbers, 10 ff. 

Kepler, 287. 
Khowarizmi, 90. 

latitude, 67, 147. 

lead of a screw, 211. 

Leibniz, 69. 

light-waves, 415. 

limacon, 437. 

limiting values, 97 ff. 

linear function, 77 ff., 101 ff., 149 ff., 

155 ff., 462 ff. 
logarithms, 41 ff. 
London Bridge, 281, 284. 
longitude, 67, 147. 

mantissa, 44. 

maximum, 398. 

mean, arithmetical, 172 ff. 

geometrical, 178. 

weighted, 173. 
measurements, 31 ff., 37. 
mercury, expansion of, 63, 85, 103. 
mid-point formula, 107. 
mil, 114, 115, 214, 215. 
minutes, 111. 
modulus, 444 ff. 
Mohammed ibn Musa al-Khowarizmi, 

90. 
multiplication, 5. 

abbreviated, 34. 

graphical, 15. 
musical scale, 414. 

Napier, 49. 

Nasir ad-Din at-Tusi, 256. 
natural logarithms, 425 ff. 
negative angles, 110. 

numbers, 6. 
Newton, 69, 203 ff., 260 ff., 287. 
Nicomedes, 437. 



normal form, of the equation of a 
line, 155 ff. 

of a plane, 470 ff. 
normal distribution curve, 428. 
du Nouy, 431. 
numbers, 1 ff. 

classification, 3. 

definition, 3. 

oblate spheroid, 476, 478. 
Oldenburg, 203. 
Omar al-Khayyam, 204. 
one-to-one correspondence, 1. 
orbit of the earth, 38, 297 ff. 
ordinate, 67. 
organ-pipes, 388. 
origin, 1, 453. 

parabola, 73, 309 ff. 
parabolic arch, 317. 

reflector, 318, 319. 
paraboloids, 481 ff. 
parallel lines, 149, 462. 

planes, 471. 
parameter, 109, 221, 281, 326, 457 ff., 

464, 468. 
Pascal, 203, 437. 
pendulum, 38, 39, 54, 100, 213, 317 ff. f 

427, 434. 
percentage error, 31, 36. 
perpendicular lines, 149, 462. 
phase-angle, 412, 416. 
piston-rod motion, 420 ff. 
point of division, 105 ff., 108, 459. 
point-slope formula, 101. 
polar-coordinates, 115 ff., 118, 435 ff. 
population statistics, 57, 65 ff., 179. 
premium, 190. 
present value, 184 ff. 
progressive computation, 195, 201 ff. 
projectiles, 87, 93, 94, 99, 100, 154, 

214 ff., 278 ff., 317. 
projecting planes, 468 ff. 
projection of vectors, 152 ff., 156 ff. 
prolate spheroid, 476, 478. 
Ptolemy, 125, 242. 
Pythagoras, 125. 

quadrants, 113. 
quadratic form, 95 ff. 

function, 87 ff. 

graphical solution, 90 ff. 

solution, 88 ff. 



522 



INDEX 



radian, 111 ff. 
radical axis, 231 ff. 

center, 233. 
railroad curves, 217 ff. 
rational numbers, 7. 
real numbers, 2. 
Recorde, 13. 

rectangular hyperbola, 327, 329. 
rectilinear generator, 491 ff. 
reflection of light, 201 ff. 
refraction of light, 261 ff. 
Regiomontanus, 257. 
regular polygons, 448, 450. 
related angles, 128 ff. 
remainder theorem, 26, 394 ff . 
resultant, 152. 

Rialto in Venice, 213, 225, 236. 
right focal chord, 292, 310, 323. 
right-handed system of axes, 454. 
roots of unity, 447 ff . 
rotation, positive and negative, 110. 
ruled surfaces, 491 ff. 

scalar line, 1. 

screw, 211. 

secant, definition, 120. 

seconds, 211. 

series, arithmetical, 166 ff. 

binomial, 193 ff. 

geometrical, 176 ff. 
significant figures, 31. 
silo, 53, 76, 100. 
simple curve, 217. 
simple harmonic motion, 409, 412. 
sine curve, 407 ff . 

definition, 117. 

law, 255 ff. 
sinking fund, 189 ff. 
sinusoid, 407 ff., 419. 
skew lines, 461. 
slope, 101, 149 ff., 397 ff. 
slope-intercept formula, 101. 
sound, 86, 99, 107, 306, 412 ff. 
specific gravity, 39. 
sphere, 458, 473, 476. 
spherical segment, 405. 
spheroids, 478. 



spiral, 211, 438. 
squaring numbers, 21, 73. 
statistics, 59 ff. 
Steinmetz, 450. 
subtraction, 4. 

geometrical, 13. 
symbols, 13. 
synthetic division, 25, 394 ff. 

tables, 40 ff., 139 ff., 495 ff. 
tabular difference, 48, 141. 
tangent, definition, 119, 125 ff. 

law, 261. 

planes, 490 ff. 

to a curve, 227, 230, 298 ff., 301 ff., 
315 ff., 490 ff. 
Tartaglia, 399. 
temperature chart, 60. 
tensor, 444. 
Thales, 266. 

transformation of coordinates, 371 ff. 
transition curves, 217, 389 ff. 
triangle, trigonometric solution, 251 

ff„ 266 ff. 
trisection of an angle, 399. 
two-point formula of a line, 101. 
tuning fork, 412« 

variable, 12. 

variation of trigonometric functions, 

126 ff. 
vector, 115, 152 ff., 444 ff. 

in space, 456. 
vectorial angle, 115. 
vibrations, 407, 412 ff. 
Viete, 13, 242. 
voice records, 413. 

water, weight, volume, etc., 61 ff., 

85, 103. 
wave lengths, 413. 
wave motion, 407. 
Wessel, 449. 
Widmann, 13. 

Zeno, 177. 



LIBRARY OF CONGRESS 




005 594 578 



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